Where a maximum can be
Both partials of $x^{2}-y^{2}$ vanish at the origin. Is the origin the top, the bottom, or neither?
In one variable a maximum inside an interval has a horizontal tangent (Fact 6.13). With two variables there are two partial derivatives, $f'_{1}$ and $f'_{2}$ (Lecture 8), one for each direction along an axis. The first question is what they must do at a top.
Let $f$ have a maximum or a minimum at an interior point $(x_{0},y_{0})$ of its domain, and let both partial derivatives exist there. Then $$ \begin{aligned} &f'_{1}(x_{0},y_{0}) = 0 \\ &\qquad\text{and}\qquad f'_{2}(x_{0},y_{0}) = 0\,. \end{aligned} $$
Hold $y$ at $y_{0}.$ Then $g(x)=f(x,y_{0})$ is a function of one variable with an interior maximum at $x_{0},$ so $g'(x_{0})=0$ by Fact 6.13 — and $g'$ is precisely $f'_{1}.$ Holding $x$ at $x_{0}$ instead gives the other equation. Nothing new is assumed: this is Lecture 6's theorem applied twice, once per variable.
The words are the same as in Lecture 6, where a critical point of a function of one variable was a point with $f'(c)=0$ (Definition 6.12). Now both partials must vanish at the same point.
Guess first. The question below is the notes' next example. Answer it before you read the solution.
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
Both partials vanish at the origin. Is the origin a maximum, a minimum, or neither?
- Why is that answer tempting? Checks one direction and stops. Everything in the option is true — f(x,0) = x² really does have a minimum at 0 — and it settles nothing about the other directions.
- Why is that answer tempting? The same mistake as A, facing the other way. Notice that A and B are each right about their own line and each wrong about the point: that is exactly the difficulty.
- Why is that answer tempting? The second derivatives would confirm it, but they are not needed: A and B already contradict each other, and two directions disagreeing is by itself enough to rule out both.
Along the $x$-axis, $f = x^{2}$ rises away from the origin. Along the $y$-axis, $f = -y^{2}$ falls away from it.
So arbitrarily close to the origin there are points where $f$ is larger and points where $f$ is smaller. It is neither a maximum nor a minimum.
Such a point is called a saddle point: the middle of a horse's back, the highest point along the ridge and the lowest point along the path across it.
Two directions that disagree
Both partials of $f(x,y)=x^{2}-y^{2}$ vanish at the origin. Is the origin a maximum, a minimum, or neither?
Try it yourself first. Then open one step at a time.
Neither. Along the $x$-axis, $f(x,0)=x^{2},$ which rises away from the origin; along the $y$-axis, $f(0,y)=-y^{2},$ which falls away from it. So arbitrarily close to the origin there are points where $f$ is larger and points where it is smaller.
Note that both of those statements are true, and that neither of them is about the origin. Two true statements about two directions do not combine into a statement about the point — which is the difficulty this whole lecture is about.
Neither: $f(t,0)=t^{2}>0$ and $f(0,t)=-t^{2}<0$ for every $t\ne0,$ while $f(0,0)=0.$
A critical point of the kind in Example 10.4 is called a saddle point: the middle of a horse's back, or a mountain pass, which really is simultaneously a highest point and a lowest one depending on which way you face. Lecture 11 defines the term precisely and gives the test that sorts critical points into the three possible kinds.
Those three kinds are all there are, and on a contour map they look nothing alike. A local maximum shows as closed loops with the numbers rising inwards; a local minimum as closed loops with the numbers rising outwards; and a saddle point as contours that cross. Often the map settles the question before any algebra is done. Every Himalayan summit is a local maximum and only Everest is the global one; the deepest point of each lake is a local minimum; and every pass between two peaks is a saddle.
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
On the square $0\le x\le1,$ $0\le y\le1,$ the function $f(x,y)=x+2y$ has its largest value $f(1,1)=3$ at the corner $(1,1).$ There $f'_{1}=1$ and $f'_{2}=2.$ What does this say about Fact 10.1?
- Why is that answer tempting? The Fact has a hypothesis: the point must be interior. A corner of the square is on its boundary, so the Fact says nothing about it, and nothing it says is contradicted.
- Why is that answer tempting? Every point of the square has $x\le1$ and $y\le1,$ so $f(x,y)=x+2y\le3=f(1,1).$ It is the maximum. Partials equal to zero are required only at an interior maximum.
- Why is that answer tempting? There is none: $f'_{1}=1$ and $f'_{2}=2$ everywhere, so no point at all is critical. The maximum is on the boundary, where Watch out 10.3 says it need not be critical.
Fact 10.1 is about interior points. At a point of the boundary you cannot move in every direction and stay in the domain, so the one-variable argument behind the Fact breaks down. Here $f$ grows in both directions everywhere, and the largest value sits in the corner where you are stopped. That is the word interior of Watch out 10.3 at work.
What the firm's rule says
A firm can take on one more worker. When should it stop hiring?
The application this chapter is built around is the firm choosing capital $K$ and labour $L.$ If output sells at price $p,$ capital costs $r$ per unit and labour $w$ per unit, profit is $$ \pi(K,L) = p\,F(K,L) - rK - wL\,, $$ and Fact 10.1 gives the two conditions $$ p\,F'_{K} = r \qquad\text{and}\qquad p\,F'_{L} = w\,. \tag{10.1} $$ Read the second one in words: hire until the extra output the last worker produces is worth exactly the wage. Above that point an extra worker brings in more than they cost, below it less. That is not a formula to memorise but a one-line marginal argument: hiring $\ell$ more workers raises revenue by about $pF'_{L}\ell$ and costs $w\ell,$ so at the optimum the two are equal. The units confirm the shape — money per unit of output times units of output per worker is money per worker, which is what a wage is — and dividing (10.1) by $p$ gives $F'_{L}=w/p{:}$ the marginal product of labour equals the real wage.
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
A firm's profit-maximising condition for labour is $$p\,F_L' = w$$
What does it say the firm should do?
- Why is that answer tempting? Average, not marginal — the recurring confusion of the whole course. $F'_{L}$ is the extra output from one more worker, not the output per worker.
- Why is that answer tempting? That is a break-even condition and it describes zero profit, not maximum profit. A firm at its profit maximum is usually making a great deal of money.
- Why is that answer tempting? Maximising output is not maximising profit. Output is largest where $F'_{L}=0$ — where the last worker adds nothing — and that is far past the point where workers stop paying for themselves.
$p\,F_L'$ is the value of the marginal product: the extra output one more worker produces, valued at the price the firm actually gets for it.
Hire while that exceeds $w;$ stop when they are equal. Dividing by $p{:}$ $$F_L' = \frac{w}{p}\,,$$ the marginal product of labour equals the real wage.
Recap The economics words
| word | what it is |
|---|---|
| marginal product of labour $F'_{L}$ | the extra output from one more worker |
| value of the marginal product $pF'_{L}$ | that extra output, valued at the price $p$ |
| average product $F/L$ | output per worker |
| real wage $w/p$ | the wage measured in units of output |
| wage bill $wL$ | what all the workers together cost |
| break-even | revenue equal to cost: zero profit, not maximum profit |
A two-product firm
A firm makes $x$ units of good A and $y$ of good B at total cost $$ \begin{aligned} C(x,y) &= 0.04x^{2} + 0.01xy + 0.01y^{2}\\ &\quad + 4x + 2y + 500\,, \end{aligned} $$ selling A at $15$ and B at $9.$ Find the candidate for maximum daily profit.
Try it yourself first. Then open one step at a time.
Profit is $$ \begin{aligned} \pi(x,y) &= 15x+9y\\ &\quad-C(x,y)\\ &= 11x + 7y\\ &\quad - 0.04x^{2} - 0.01xy\\ &\quad - 0.01y^{2} - 500\,, \end{aligned} $$ so the first-order conditions are $$ \begin{aligned} \pi'_{1} &= 11 - 0.08x - 0.01y\\ &= 0\,,\\ \pi'_{2} &= 7 - 0.01x - 0.02y\\ &= 0\,. \end{aligned} $$
These are linear, because $\pi$ is quadratic — which is why so many models in economics are quadratic. The second gives $x = 700-2y;$ substituting into the first gives $56 - 0.16y + 0.01y = 11,$ so $y=300$ and $x=100,$ with $\pi(100,300) = 1100.$
And now the sentence that matters: we have found a candidate and we have not finished. Nothing so far rules out a minimum or a saddle. Example 10.15 finishes it.
The only candidate is $(x,y)=(100,300),$ with daily profit $\pi(100,300)=1100.$ Whether it is the maximum is not yet decided.
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
Find the critical point of $$\pi(x,y)=10x+8y-x^{2}-xy-y^{2}\,.$$
- Why is that answer tempting? Leaves out the cross term. The partial of $-xy$ with respect to $x$ is $-y,$ so $\pi'_{1}=10-2x-y,$ not $10-2x.$ The two equations are tied together, as in Example 10.6.
- Why is that answer tempting? The first-order conditions are necessary, not sufficient (Watch out 10.3): they produce candidates. Showing that $(4,2)$ is the maximum takes the concavity of the next sections.
- Why is that answer tempting? The values of $x$ and $y$ swapped. Check by substituting: at $(2,4),$ $\pi'_{1}=10-4-4=2\ne0.$
The first-order conditions are $$ \begin{aligned} \pi'_{1}&=10-2x-y=0\,,\\ \pi'_{2}&=8-x-2y=0\,. \end{aligned} $$ The first gives $y=10-2x;$ then $8-x-20+4x=0,$ so $x=4$ and $y=2.$ That is a candidate and nothing more, until something else rules out a minimum or a saddle.
Two directions are not enough
$3xy-x^{2}-y^{2}$ curves downwards along the $x$-axis and along the $y$-axis. Is the origin its top?
In one variable, $f'(c)=0$ together with $f''(c)<0$ settles the question. The obvious generalisation would be that both partials vanish, $f''_{11}<0$ and $f''_{22}<0.$ It is false, and one function shows it.
Recap The notation for second-order partials
Lecture 8 differentiated twice. For $f(x,y)$:
- $f''_{11}$ is the partial of $f'_{1}$ with respect to $x,$ the second derivative along the $x$-direction;
- $f''_{22}$ is the partial of $f'_{2}$ with respect to $y,$ the second derivative along the $y$-direction;
- $f''_{12}=f''_{21}$ is the partial of $f'_{1}$ with respect to $y,$ the cross-partial.
For $f=3xy-x^{2}-y^{2}{:}$ $f'_{1}=3y-2x$ and $f'_{2}=3x-2y,$ so $f''_{11}=-2,$ $f''_{22}=-2$ and $f''_{12}=3.$
Guess first. Answer the question below before you read the example after it.
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
For $f(x,y) = 3xy - x^{2} - y^{2}$, both partials vanish at the origin and $$f_{11}'' = f_{22}'' = -2 < 0$$ so $f$ has a maximum along the $x$-axis and along the $y$-axis.
Does $f$ have a maximum at the origin?
- Why is that answer tempting? The guess this whole section exists to refute. Two directions out of infinitely many, and no argument connecting them.
- Why is that answer tempting? Weakens the claim instead of testing it. Local or global, it is not a maximum in any neighbourhood, however small.
- Why is that answer tempting? Half right, and it stops one step too soon. The cross-partial is indeed what settles it — but the function is in front of you, so you can compute f″₁₂ = 3, or substitute y = x and be finished in one line.
Along the line $y = x$ the origin is a minimum. So arbitrarily close to it there are points above and points below: a saddle.
Both direct second derivatives said "maximum", and both were right about their own line. What they missed is the cross term $f_{12}'' = 3$, which measures how much the surface twists — and here it twists enough to open a valley along the diagonal.
Concave along both axes, and not a maximum
Let $f(x,y) = 3xy - x^{2} - y^{2}.$ Both partials vanish at the origin, and $$ f''_{11} = f''_{22} = -2 < 0\,, $$ so $f$ is concave along the $x$-axis and along the $y$-axis. Is the origin a maximum?
Try it yourself first. Then open one step at a time.
No. Walk the diagonal instead: $$ \begin{aligned} f(t,t) &= 3t^{2} - t^{2} - t^{2}\\ &= t^{2}\,, \end{aligned} $$ which has a minimum at $t=0.$ So along that one line the origin is the bottom of a valley, and the origin is a saddle point.
What the two direct second derivatives missed is the cross term $f''_{12}=3,$ which measures how much the surface twists. Here it twists enough to open a valley along a direction neither of them looked at.
No. Along the line $y=x$ the function is $t^{2},$ larger than $f(0,0)=0$ for every $t\ne0{:}$ the origin is a saddle point.
Every direction at once
Four functions, all with both partials zero at the origin. Left is the contour map with a ray through the origin, in the middle $f$ along that ray, and on the right how $f$ curves in every direction at once: outside the dashed circle the ray curves up, inside it curves down.
- With the first function, “a maximum”, press spin. The curve on the right stays inside the circle: every direction curves down.
- Press the button with the function's name until it reads “an obvious saddle” ($x^{2}-y^{2}$), and spin again. The curve crosses the circle: some directions curve up, some down.
- Go on to “the sneaky one” ($3xy-x^{2}-y^{2}$). Set the direction to $0°$ and to $90°{:}$ both look like a maximum. Now set it to $45°,$ the diagonal: a valley.
- Under the picture, $A,$ $B$ and $C$ are $f''_{11},$ $f''_{12}$ and $f''_{22},$ and $AC-B^{2}$ is the left-hand side of the third condition in the next section. Watch its sign as you change function: it is negative exactly when the curve crosses the circle.
Fixing this needs a condition that controls every direction at once, and the right one is a two-variable version of the concavity of Lecture 6. That is what the next subsection supplies.
Concavity in two variables
Which condition, checked once, controls how $f$ curves in every direction at the same time?
Lecture 6 defined a concave function of one variable by the chord condition, and proved that an interior stationary point of a concave function is a global maximum. Everything here is the same statement with pairs in place of numbers — but one preliminary is needed that one variable did not require.
A disc, a rectangle, a half-plane and the whole plane are convex; a ring, a crescent and any set with a hole or a notch are not. The reason the notion is needed at all is that the definition below evaluates $f$ at points on the segment between two points of the domain, and unless the domain is convex those points may not be in it, leaving the inequality meaningless.
(A convex set is a property of a set; a convex function, as in Lecture 6, is a property of a function. The two words are the same and the ideas are different.)
Let $f$ have continuous second-order partial derivatives on a convex set $S$. Then $f$ is concave on $S$ if and only if, at every point of $S$, $$ \begin{aligned} &f''_{11}\le0\,, \\ &f''_{22}\le0\,, \\ &f''_{11}f''_{22} - \bigl(f''_{12}\bigr)^{2} \ge 0\,. \end{aligned} \tag{10.2} $$ Reversing the first two inequalities, and keeping the third, characterises convexity.
The first two conditions are the ones anybody would guess, and Example 10.7 showed they are not enough on their own. The third is the new one and it does all the work: it caps how far the surface may twist. For $3xy-x^{2}-y^{2}$ it fails, and by exactly the right amount, since $(-2)(-2)-3^{2} = -5 < 0.$
Guess first. Work the question below out from (10.2) before you read on.
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
At a critical point, $f_{11}'' = -1$ and $f_{22}'' = -4$.
For the concavity test to certify a maximum, what must be true of the cross-partial $f_{12}''$?
- Why is that answer tempting? Precisely the belief Example 10.7 refutes. f″₁₂ = 5 would give (−1)(−4) − 25 = −21, and the point would be a saddle.
- Why is that answer tempting? Confuses the sign of the cross-partial with its size. Only its square appears in the test, so −5 fails exactly as badly as +5.
- Why is that answer tempting? Too strong. A surface may twist a little and still be concave; it is only a large twist that breaks the test.
The third condition is $$f_{11}''\,f_{22}'' - \left(f_{12}''\right)^{2} \ge 0$$ so $(-1)(-4) - (f_{12}'')^{2} \ge 0$, that is $(f_{12}'')^{2} \le 4$, that is $|f_{12}''| \le 2$.
Only the square of the cross-partial appears. Its sign is irrelevant; its size is everything. The condition is a cap on how far the surface may twist before some diagonal direction curves the wrong way.
What the third condition is for
At a critical point, $f''_{11}=-1$ and $f''_{22}=-4.$ What must be true of $f''_{12}$ for (10.2) to certify a maximum?
Try it yourself first. Then open one step at a time.
The first two conditions already hold. The third requires $$ (-1)(-4) - \bigl(f''_{12}\bigr)^{2} \ge 0\,, $$ $$ \text{that is}\qquad \bigl|f''_{12}\bigr| \le 2\,. $$
So the surface is allowed to twist, but only so far. A cross-partial of $5$ would give $4-25=-21$ and the point would be a saddle, exactly as in Example 10.7.
$|f''_{12}|\le2,$ that is $-2\le f''_{12}\le2;$ the sign does not matter.
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
Is $f(x,y)=-x^{2}+4xy-5y^{2}+2x$ concave on the whole plane?
- Why is that answer tempting? The sign of the cross-partial is irrelevant: only $\bigl(f''_{12}\bigr)^{2}=16$ enters (10.2) (Watch out 10.12).
- Why is that answer tempting? The third condition compares $\bigl(f''_{12}\bigr)^{2}$ with the product $f''_{11}f''_{22},$ not with one of them: here $16$ against $(-2)(-10)=20.$
- Why is that answer tempting? The second-order partials of a quadratic are constants, so the three conditions hold at every point of the plane or at none.
$f''_{11}=-2,$ $f''_{22}=-10$ and $f''_{12}=4,$ constants. The first two conditions hold, and $$ \begin{aligned} &(-2)(-10)-4^{2}\\ &=20-16=4\ge0\,. \end{aligned} $$ The plane is a convex set, so by Fact 10.11 $f$ is concave on it.
Concavity decides, globally
Example 10.6 found the candidate $(100,300).$ What does it take to call it the best plan?
Concavity is worth checking because of what it buys: the same conclusion as Lecture 6's Fact 6.20, with a plane in place of a line.
If $f$ is concave on a convex set $S$ and $(x_{0},y_{0})$ is an interior critical point, then $(x_{0},y_{0})$ is a maximum point for $f$ over the whole of $S$. If $f$ is convex, such a point is a minimum over the whole of $S$.
For a concave function the tangent plane at any point lies on or above the graph everywhere. At a critical point both partials vanish, so that tangent plane is horizontal, at height $f(x_{0},y_{0}).$ Hence the whole graph lies at or below that height, which is what the claim says. It is Lecture 6's one-variable argument with a plane in place of a line.
The two-product firm, finished
Finish Example 10.6: show that $(100,300)$ really is the profit-maximising plan.
Try it yourself first. Then open one step at a time.
The second-order partials of $$ \begin{aligned} \pi &= 11x+7y-0.04x^{2}\\ &\quad-0.01xy-0.01y^{2}-500 \end{aligned} $$ are constants: $$ \begin{aligned} \pi''_{11} &= -0.08\,,\\ \pi''_{22} &= -0.02\,,\\ \pi''_{12} &= -0.01\,, \end{aligned} $$ so the first two conditions of (10.2) hold, and $$ \begin{aligned} &\pi''_{11}\pi''_{22} - \bigl(\pi''_{12}\bigr)^{2}\\ &= (-0.08)(-0.02) - (-0.01)^{2}\\ &= 0.0016 - 0.0001\\ &= 0.0015 > 0\,. \end{aligned} $$
All three hold everywhere — they do not depend on $x$ or $y$ at all — and the domain is the whole plane, which is convex. So $\pi$ is concave, and by Fact 10.13 the critical point $(100,300)$ is a global maximum. Daily profit is $1100,$ and now that is a conclusion rather than a candidate.
$\pi$ is concave on the whole plane, so $(100,300)$ is the global maximum point: the best plan, with daily profit $1100.$
For a minimum the same steps run with the first two inequalities of (10.2) reversed: a convex function on a convex set has its global minimum at every interior critical point.
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
Let $f(x,y)=2xy-x^{2}-y^{2}.$ Both partials vanish at the origin, and $f''_{11}=f''_{22}=-2,$ $f''_{12}=2.$ What is the origin?
- Why is that answer tempting? (10.2) asks for $\ge0,$ and $(-2)(-2)-2^{2}=0$ satisfies it. Zero is not a failure.
- Why is that answer tempting? Along the diagonal $f(t,t)=0,$ and along $y=-x$ it is $-4t^{2}.$ No line through the origin goes above $0.$ Concavity controls every direction at once.
- Why is that answer tempting? The conditions hold at every point, so Fact 10.13 gives a global verdict now. The local test is needed only when they fail somewhere.
The three conditions hold at every point: $-2\le0,$ $-2\le0,$ and $(-2)(-2)-2^{2}=0\ge0.$ So $f$ is concave on the plane, and by Fact 10.13 the critical point $(0,0)$ is a global maximum point. Indeed $f=-(x-y)^{2}\le0=f(0,0).$ (Every point of the line $y=x$ is a maximum point too: a maximum need not be unique.)
What a constraint does
The best plan totals $400$ units. If at most $320$ are allowed, where is the best plan now?
Real problems come with constraints, and a constraint moves the answer.
Guess first. The firm of Example 10.6 is now held to a total of $320$ units. Choose a method before you read on.
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
A firm's profit is maximised at $(x, y) = (100, 300)$. It is now restricted to $x + y = 320$.
What is the right way to find the new best plan?
- Why is that answer tempting? Assumes the best mix does not change when the total does. Nothing guarantees that, and here it is false: the answer is (90, 230), not (80, 240).
- Why is that answer tempting? Three equations in two unknowns, and in general no solution at all. At a constrained optimum the partials are not zero — which is exactly why the constraint changes the answer.
- Why is that answer tempting? Distance in the (x, y) plane has no meaning here. Profit contours are not circles, and the nearest point is not the best point.
Substituting turns a constrained two-variable problem into an unconstrained one-variable one: $$ \begin{aligned} \hat\pi(x) &= \pi(x,\,320-x)\\ &= 7.2x - 0.04x^{2} + 716 \end{aligned} $$ so $\hat\pi'(x) = 7.2 - 0.08x = 0$ gives $x = 90,$ $y = 230,$ and $\hat\pi'' < 0$ confirms it.
Profit falls from $1100$ to $1040$ — and the answer is where the line touches a contour, as in Lecture 9.
When the constraint can be solved for one of the variables there is a cheap way through: substitute, and a constrained two-variable problem becomes an unconstrained one-variable problem of the kind Lecture 6 already handles.
A cap on total output
The firm of Example 10.6 is now restricted to total output $x+y=320.$ Find the best available plan.
Try it yourself first. Then open one step at a time.
The unconstrained answer $(100,300)$ totals $400$ and is no longer available. Substituting $y=320-x$ into $\pi$ and collecting terms, $$ \hat\pi(x) = 7.2x - 0.04x^{2} + 716\,, $$ so $\hat\pi'(x) = 7.2-0.08x = 0$ gives $x=90,$ and $\hat\pi''=-0.08<0$ confirms a maximum. Hence $x=90,$ $y=230,$ and profit falls from $1100$ to $\hat\pi(90)=1040.$
Two things are worth noticing. First, the best mix changed: simply scaling $(100,300)$ down to the new total would give $(80,240)$ and a profit of $1036$ — only four less, which is why the error goes unnoticed, and still the wrong answer. Second, geometrically the solution is where the constraint line touches a contour of $\pi,$ which is exactly the tangency argument of Lecture 9.
$x=90,$ $y=230,$ with profit $1040$ (against $1100$ without the cap).
The line in Example 10.20 runs on for ever, but output cannot be negative. The plans that are really available are the segment $x+y=320,$ $x\ge0,$ $y\ge0,$ and a segment has two ends. On a closed interval Lecture 6 compared the stationary points with the endpoints (Method 6.15); after the substitution that is exactly the situation here.
Largest and smallest on a segment, endpoints included
This example is the app's; it joins Method 10.19 to Method 6.15 of Lecture 6, and nothing in it is new. The firm of Example 10.6 must produce exactly $c$ units in total, with $x\ge0$ and $y\ge0.$
(a) For $c=320,$ find the largest and the smallest value of $\pi$ on the segment.
(b) For $c=40,$ find the best plan.
Try it yourself first. Then open one step at a time.
(a) One variable on a closed interval. On the segment $y=320-x$ with $0\le x\le320,$ and by Example 10.20 $$ \hat\pi(x)=7.2x-0.04x^{2}+716\,. $$ $\hat\pi$ is continuous on the closed and bounded interval $[0,320],$ so by the extreme value theorem (Fact 6.9) both a largest and a smallest value exist.
(a) The candidates. By Method 6.15: the stationary point $x=90$ of $\hat\pi,$ which lies inside $(0,320),$ and the two endpoints. $$ \begin{aligned} \hat\pi(90)&=1040\,,\\ \hat\pi(0)&=716\,,\\ \hat\pi(320)&=2304-4096+716\\ &=-1076\,. \end{aligned} $$
(a) Compare. The largest value is $1040$ at $(90,230);$ the smallest is $-1076$ at the endpoint $(320,0),$ where the firm makes only good A. The stationary point gave the maximum, and an endpoint the minimum.
(b) A tight cap. For $c=40$ put $y=40-x$ with $0\le x\le40.$ Collecting terms as before, $$ \begin{aligned} \hat\pi(x)&=\pi(x,40-x)\\ &=-0.04x^{2}+4.4x-236\,, \end{aligned} $$ so $\hat\pi'(x)=4.4-0.08x,$ which is zero at $x=55.$ That is outside $[0,40]{:}$ on the whole interval $\hat\pi'(x)\ge4.4-3.2=1.2>0,$ so $\hat\pi$ is increasing there.
(b) The endpoint wins. The largest value is at the endpoint $x=40{:}$ $\hat\pi(40)=-64+176-236=-124,$ against $\hat\pi(0)=-236.$ The best plan is $(40,0),$ only good A. The firm still makes a loss — $40$ units cannot cover the fixed cost of $500$ — but a smaller loss than with any other mix.
(a) Largest $1040$ at $(90,230),$ smallest $-1076$ at the endpoint $(320,0).$
(b) The endpoint $(40,0),$ with $\pi=-124{:}$ the stationary point $x=55$ of $\hat\pi$ lies outside the segment.
Pick an answer before reading on. Getting it wrong is fine — that is how you find out what you believe.
What is the largest value of $f(x,y)=x^{2}+y^{2}$ on the segment $x+y=4,$ $x\ge0,$ $y\ge0$?
- Why is that answer tempting? $\hat f(x)=x^{2}+(4-x)^{2}=2x^{2}-8x+16$ has its stationary point at $x=2,$ but $\hat f''=4>0{:}$ that is the smallest value. A stationary point is a candidate, and the endpoints are candidates too (Method 6.15).
- Why is that answer tempting? The segment is closed and bounded: after the substitution $\hat f$ is continuous on $[0,4],$ so by Fact 6.9 a largest value exists.
- Why is that answer tempting? $(4,4)$ is not on the segment: $4+4\ne4.$ The constraint ties $y$ to $x.$
Put $y=4-x$ with $0\le x\le4{:}$ $$ \begin{aligned} \hat f(x)&=x^{2}+(4-x)^{2}\\ &=2x^{2}-8x+16\,. \end{aligned} $$ The candidates are the stationary point $x=2$ and the endpoints: $\hat f(2)=8,$ $\hat f(0)=\hat f(4)=16.$ The largest value is $16,$ at both endpoints; the smallest is $8,$ at $(2,2).$
Lecture 11 turns that tangency picture into a method that works whether or not the constraint can be solved.
What you should take away
- At an interior extreme point both partials vanish — Lecture 6's theorem applied once per variable. Such a point is a critical point, and the condition is necessary and not sufficient for the sixth time this term.
- There are exactly three kinds of critical point: local maximum, local minimum and saddle. On a contour map they are loops rising inwards, loops rising outwards, and crossing contours.
- The firm's rule $pF'_{L}=w$ says: hire until the last worker's extra output is worth the wage. Dividing by $p,$ marginal product equals the real wage.
- Two directions are not enough. $f=3xy-x^{2}-y^{2}$ has $f''_{11}=f''_{22}=-2$ and $f(t,t)=t^{2}{:}$ concave along both axes and a valley along the diagonal.
- A set is convex if it contains the segment between any two of its points, and $f$ is concave on such a set if chords lie below the graph. That holds if and only if $f''_{11}\le0,$ $f''_{22}\le0$ and $f''_{11}f''_{22}-(f''_{12})^{2}\ge0$ throughout. Only the square of the cross-partial appears, so its sign never matters.
- Concavity throughout a convex domain turns every interior critical point into a global maximum, because the tangent plane lies above the graph and at a critical point it is horizontal. The conditions must hold everywhere; at a single point the verdict is only local.
- A constraint moves the answer, and at a constrained optimum the partials are not zero. If the constraint can be solved for one variable, substitute and the problem becomes one-variable.
Think further
Questions without computation. Think on your own or with a classmate before opening the thought.
Discussion question 1 — two true statements that settle nothing
Both partials of $f(x,y)=x^{2}-y^{2}$ vanish at the origin. Is it a maximum, a minimum, or neither? Two of the wrong answers are true statements — which, and why do they settle nothing?
One way to think about it
Neither (Example 10.4). The two wrong answers built on true statements are “a minimum, because $f(x,0)=x^{2}$ has a minimum at $0$” and “a maximum, because $f(0,y)=-y^{2}$ has a maximum at $0$”. Each is right about its own line. But a maximum must beat every point near the origin, and a minimum must lie below every one, and each line is only one of infinitely many directions. Here the two lines even contradict each other: there are points with $f>0$ and points with $f<0$ arbitrarily close to the origin, where $f=0.$
Discussion question 2 — what the hiring rule says
A firm's condition for labour is $pF'_{L}=w.$ What does it say?
(A) hire until the wage equals average output per worker;
(B) hire until the last worker's extra output is worth the wage;
(C) hire until revenue equals the wage bill;
(D) hire until output is as large as possible.
One way to think about it
(B). $F'_{L}$ is the extra output of one more worker and $p$ turns it into money, so $pF'_{L}$ is what the last worker brings in; the rule stops hiring when that equals what the worker costs, $w.$ (A) uses the average product, not the marginal one (Watch out 10.5). (C) is a break-even condition: it describes zero profit, not maximum profit. (D) is where $F'_{L}=0,$ far past the point where workers stop paying for themselves.
Discussion question 3 — no derivatives needed
For $f=3xy-x^{2}-y^{2}$ both direct second derivatives are $-2.$ Without differentiating anything, show that the origin is not a maximum.
One way to think about it
Substitute $y=x{:}$ $f(t,t)=3t^{2}-t^{2}-t^{2}=t^{2}.$ For every $t\ne0$ this is larger than $f(0,0)=0,$ and $(t,t)$ is as close to the origin as we like. So no neighbourhood of the origin, however small, has its largest value there: the origin is not a maximum (Example 10.7).
Discussion question 4 — a cap on the twist
At a critical point $f''_{11}=-1$ and $f''_{22}=-4.$ Which values of $f''_{12}$ let the test certify a maximum? Why does the sign of $f''_{12}$ not matter?
One way to think about it
The third condition of (10.2) is $(-1)(-4)-\bigl(f''_{12}\bigr)^{2}\ge0,$ that is $\bigl(f''_{12}\bigr)^{2}\le4,$ that is $-2\le f''_{12}\le2$ (Example 10.16). The sign does not matter because only the square of $f''_{12}$ appears: $f''_{12}=-3$ and $f''_{12}=3$ both give $4-9=-5<0$ (Watch out 10.12).
Discussion question 5 — why the domain must be convex
Why must the domain be convex in Definition 10.10? Give a set for which the inequality would not even make sense.
One way to think about it
The inequality evaluates $f$ at $\bigl(\lambda x_{1}+(1-\lambda)x_{2},\,\lambda y_{1}+(1-\lambda)y_{2}\bigr),$ a point on the segment between two points of the domain. If that point is outside the domain, $f$ has no value there and the inequality says nothing. Take the ring $1\le x^{2}+y^{2}\le4.$ The points $(-1.5,0)$ and $(1.5,0)$ lie in it, but the midpoint $(0,0)$ (with $\lambda=\tfrac12$) does not. A function defined only on the ring, such as $\ln(x^{2}+y^{2}-\tfrac12),$ has no value there to compare.
Discussion question 6 — the word that would have failed
Example 10.15 checked (10.2) and found the conditions held everywhere. Which word in Fact 10.13 would have failed if they had held only at $(100,300)?$
One way to think about it
Whole — “a maximum point for $f$ over the whole of $S$”. The Fact needs $f$ concave on $S,$ which by Fact 10.11 means the three conditions at every point of $S.$ Holding at the single point $(100,300)$ they would say something about points near it only: a local verdict, which is Lecture 11's (Watch out 10.14).
Discussion question 7 — why it is not a method
In Example 10.20, why is “set both partials of $\pi$ to zero and impose $x+y=320$” not a method?
One way to think about it
Because it asks three equations to hold in two unknowns (Watch out 10.18). Both partials vanish only at $(100,300),$ and $100+300=400\ne320,$ so no point satisfies all three. At the best available plan $(90,230)$ the partials are not zero: $$ \begin{aligned} \pi'_{1}&=11-7.2-2.3=1.5\,,\\ \pi'_{2}&=7-0.9-4.6=1.5\,. \end{aligned} $$ They are equal, which says that moving one unit from one good to the other changes profit by nothing to first order — the tangency of Lecture 9, which Lecture 11 turns into a method.