10 Problems
Two directions are not enough
The problems follow the sections of the theory. Those that are checked at once get new numbers each time you press New variant: repeat them until the method runs by itself. Those with a solution are for paper. Write the whole solution before you open it, then compare line by line. Last come problems in the form of the exam, with points and a marking guide.
Typing an answer. Every answer checked here is a number. Write fractions as 7/2 or decimals as 3.5, and a negative number with a minus sign, -12.
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Where a maximum can be
Read the theory: 1. Where a maximum can be →Let $f(x,y)=xy.$
(a) Find every critical point of $f.$
(b) Compute $f(t,t)$ and $f(t,-t).$ Is the critical point a maximum, a minimum, or neither?
(c) On the rectangle $0\le x\le2,$ $0\le y\le3,$ the largest value of $f$ is $f(2,3)=6.$ Both partials are non-zero there. Why does this not contradict Fact 10.1?
Hint
In (b), compare with $f(0,0)$ for small $t\ne0.$ In (c), read the hypothesis of the Fact, and Watch out 10.3.
(a) The first-order conditions. $f'_{1}=y$ and $f'_{2}=x.$ Both vanish only at $(0,0),$ the one critical point.
(b) Two lines. $f(t,t)=t^{2}>0$ and $f(t,-t)=-t^{2}<0$ for every $t\ne0,$ while $f(0,0)=0.$ Arbitrarily close to the origin there are larger and smaller values: neither a maximum nor a minimum. (It is a saddle point, the kind Lecture 11 names.)
(c) The maximum on the rectangle. For $0\le x\le2$ and $0\le y\le3$ we have $xy\le2\cdot3=6,$ so $(2,3)$ is the maximum point. There $f'_{1}=y=3$ and $f'_{2}=x=2.$
No contradiction. Fact 10.1 is about an interior maximum. The corner $(2,3)$ lies on the boundary of the rectangle: from there one cannot move in every direction and stay in the rectangle. On the boundary a maximum need not be critical at all (Watch out 10.3).
Answer
(a) Only $(0,0).$
(b) $f(t,t)=t^{2},$ $f(t,-t)=-t^{2}{:}$ neither.
(c) The maximum is at a corner, not at an interior point, and Fact 10.1 is only about interior points.
How did it go?
Stuck? Read 1. Where a maximum can be →
What the firm's rule says
Read the theory: 2. What the firm's rule says →A bakery sells bread at $p=4$ a loaf and pays a wage of $w=100$ a day. With its present staff, one more baker would add about $F'_{L}=30$ loaves a day, and the bakers produce on average $50$ loaves a day each.
(a) Should the bakery hire one more baker? About how much would daily profit change?
(b) A colleague says: “each baker brings in $4\cdot50=200,$ twice the wage, so the bakery should keep hiring.” What is wrong with the argument?
(c) What is the real wage here, and what is $F'_{L}$ at the best number of bakers?
Hint
Compare the value of the marginal product $pF'_{L}$ with the wage (the text before Watch out 10.5).
(a) The marginal argument. One more baker brings in about $pF'_{L}=4\cdot30=120$ a day and costs $w=100.$ So hiring raises daily profit by about $120-100=20.$ Hire.
(b) Average against marginal. $200$ is the value of the average product. The rule (10.1) is about the last baker hired (Watch out 10.5). As more bakers share the ovens, each new one adds less, and hiring should stop when the next one adds only $100$ worth of bread, however large the average still is.
(c) The real wage is $w/p=100/4=25$ loaves a day. Dividing $pF'_{L}=w$ by $p,$ at the best number of bakers $F'_{L}=w/p=25{:}$ the last baker adds $25$ loaves a day.
Answer
(a) Yes: $pF'_{L}=120>100,$ about $20$ more profit a day.
(b) It uses the average product; the rule is about the last baker.
(c) $w/p=25$ loaves a day, and $F'_{L}=25$ at the optimum.
How did it go?
Stuck? Read 2. What the firm's rule says →
Two directions are not enough
Read the theory: 3. Two directions are not enough →Let $f(x,y)=x^{2}+3xy+y^{2}.$
(a) Show that both partials vanish at the origin, and that $f''_{11}>0$ and $f''_{22}>0.$
(b) Show, without any further derivative, that the origin is not a minimum point.
(c) Compute $f''_{11}f''_{22}-\bigl(f''_{12}\bigr)^{2}.$
Hint
Turn Example 10.7 upside down: this time look for a line along which $f$ falls away from the origin. Try $y=-x.$
(a) $f'_{1}=2x+3y$ and $f'_{2}=3x+2y,$ both $0$ at $(0,0).$ $f''_{11}=2>0$ and $f''_{22}=2>0{:}$ along each axis $f$ is a parabola opening upwards, $f(x,0)=x^{2},$ $f(0,y)=y^{2}.$
(b) Walk the other diagonal. $$ \begin{aligned} f(t,-t)&=t^{2}-3t^{2}+t^{2}\\ &=-t^{2}<0=f(0,0) \end{aligned} $$ for every $t\ne0.$ So arbitrarily close to the origin $f$ takes values below $f(0,0){:}$ the origin is not a minimum point. (Along $y=x,$ $f(t,t)=5t^{2}>0,$ so it is no maximum either.)
(c) $f''_{12}=3,$ so $f''_{11}f''_{22}-\bigl(f''_{12}\bigr)^{2}=4-9=-5<0.$ The third condition of (10.2) fails: the twist is too large, exactly as in Example 10.7 with the signs reversed.
Answer
(a) $f'_{1}=2x+3y,$ $f'_{2}=3x+2y;$ $f''_{11}=f''_{22}=2.$
(b) $f(t,-t)=-t^{2}<0$ for $t\ne0.$
(c) $4-9=-5.$
How did it go?
Stuck? Read 3. Two directions are not enough →
Concavity in two variables
Read the theory: 4. Concavity in two variables →Decide whether each function is concave, convex, or neither on the whole plane.
(a) $f(x,y)=-3x^{2}+2xy-y^{2}+5y$
(b) $g(x,y)=x^{2}+4xy+3y^{2}$
(c) $h(x,y)=2x^{2}-3xy+2y^{2}-x$
Hint
The second-order partials of a quadratic are constants, so (10.2) is checked once and holds everywhere or nowhere. For convexity, reverse the first two inequalities and keep the third (Fact 10.11).
(a) $f''_{11}=-6,$ $f''_{22}=-2,$ $f''_{12}=2.$ Both direct ones are $\le0,$ and $(-6)(-2)-2^{2}=12-4=8\ge0.$ The plane is convex, so $f$ is concave.
(b) $g''_{11}=2,$ $g''_{22}=6,$ $g''_{12}=4.$ The third condition fails: $2\cdot6-4^{2}=12-16=-4<0.$ Both concavity and convexity need it, so $g$ is neither. (Along $y=-x,$ $g(t,-t)=t^{2}-4t^{2}+3t^{2}=0,$ and along $y=-\tfrac12x,$ $g(2t,-t)=4t^{2}-8t^{2}+3t^{2}=-t^{2}{:}$ the graph bends down there and up along the axes.)
(c) $h''_{11}=4,$ $h''_{22}=4,$ $h''_{12}=-3.$ Both direct ones are $\ge0,$ and $4\cdot4-(-3)^{2}=16-9=7\ge0.$ So $h$ is convex. The minus sign of $h''_{12}$ plays no part (Watch out 10.12).
Answer
(a) Concave. (b) Neither ($12-16<0$). (c) Convex.
How did it go?
Stuck? Read 4. Concavity in two variables →
Concavity decides, globally
Read the theory: 5. Concavity decides, globally →Let $$f(x,y)=4\ln x+2\ln y-2x-y$$ for $x>0,$ $y>0.$
(a) Find the critical point of $f.$
(b) Show that it is a global maximum point of $f$ on its domain, and give the maximum value.
Hint
$(\ln x)'=1/x.$ For (b), Method 10.17: the domain $x>0,$ $y>0$ is a convex set, and (10.2) must hold at every point of it.
(a) The first-order conditions. $$ \begin{aligned} f'_{1}&=\frac4x-2=0\,,\\ f'_{2}&=\frac2y-1=0\,, \end{aligned} $$ so $x=2$ and $y=2.$ The only critical point is $(2,2).$
(b) The domain is convex. The quadrant $x>0,$ $y>0$ contains the segment between any two of its points: both coordinates of a point on it lie between positive numbers.
(b) The three conditions everywhere. For all $x>0,$ $y>0,$ $$ \begin{aligned} f''_{11}&=-\frac{4}{x^{2}}<0\,,\\ f''_{22}&=-\frac{2}{y^{2}}<0\,,\\ f''_{12}&=0\,, \end{aligned} $$ and $f''_{11}f''_{22}-0^{2}=\dfrac{8}{x^{2}y^{2}}>0.$ So $f$ is concave on the quadrant (Fact 10.11), and by Fact 10.13 the interior critical point $(2,2)$ is a maximum point over the whole domain.
(b) The value. $$ \begin{aligned} f(2,2)&=4\ln2+2\ln2-4-2\\ &=6\ln2-6\approx-1.84\,. \end{aligned} $$
Answer
(a) $(2,2).$
(b) $f$ is concave on the convex set $x>0,$ $y>0,$ so $(2,2)$ is the global maximum point; the maximum value is $6\ln2-6\approx-1.84.$
How did it go?
Stuck? Read 5. Concavity decides, globally →
What a constraint does
Read the theory: 6. What a constraint does →A firm's profit from $x$ units of one good and $y$ of another is $$\pi(x,y)=10x+12y-x^{2}-y^{2}\,.$$
(a) Find the plan that maximises profit with no constraint, and show that it is the global maximum.
(b) The firm must now produce exactly $8$ units in total, with $x\ge0,$ $y\ge0.$ Find the largest and the smallest value of $\pi$ on this segment.
(c) Compute $\pi'_{1}$ and $\pi'_{2}$ at the best plan of (b). Are they zero? Explain why they are equal.
Hint
(a) Method 10.17. (b) Method 10.19, then compare with both endpoints, as in the theory's example on a segment. (c) Differentiate $\hat\pi(x)=\pi(x,8-x)$ by the chain rule of Lecture 9.
(a) The candidate. $\pi'_{1}=10-2x=0$ and $\pi'_{2}=12-2y=0$ give $(5,6),$ with $\pi(5,6)=50+72-25-36=61.$
(a) Global. $\pi''_{11}=-2,$ $\pi''_{22}=-2,$ $\pi''_{12}=0,$ and $(-2)(-2)-0=4\ge0,$ at every point of the plane. So $\pi$ is concave and $(5,6)$ is the global maximum point (Fact 10.13). It totals $11$ units.
(b) Substitute. On the segment $y=8-x,$ $0\le x\le8{:}$ $$ \begin{aligned} \hat\pi(x)&=10x+12(8-x)\\ &\quad-x^{2}-(8-x)^{2}\\ &=-2x^{2}+14x+32\,. \end{aligned} $$ $\hat\pi$ is continuous on $[0,8],$ so both extreme values exist (Fact 6.9).
(b) The candidates. $\hat\pi'(x)=-4x+14=0$ at $x=3.5,$ inside $[0,8].$ $$ \begin{aligned} \hat\pi(3.5)&=-24.5+49+32\\ &=56.5\,,\\ \hat\pi(0)&=32\,,\\ \hat\pi(8)&=-128+112+32\\ &=16\,. \end{aligned} $$ Largest $56.5$ at $(3.5,4.5);$ smallest $16$ at the endpoint $(8,0).$
(c) Not zero. At $(3.5,4.5),$ $\pi'_{1}=10-7=3$ and $\pi'_{2}=12-9=3$ (Watch out 10.18).
(c) Why equal. By the chain rule along $y=8-x,$ $$ \begin{aligned} \hat\pi'(x)&=\pi'_{1}+\pi'_{2}\cdot(-1)\\ &=\pi'_{1}-\pi'_{2}\,. \end{aligned} $$ At the best interior point $\hat\pi'(x)=0,$ so $\pi'_{1}=\pi'_{2}.$ In words: moving one unit from one good to the other gains $\pi'_{1}$ and loses $\pi'_{2},$ and at the best plan on the line the two balance.
Answer
(a) $(5,6),$ $\pi=61;$ global because $\pi$ is concave.
(b) Largest $56.5$ at $(3.5,4.5),$ smallest $16$ at $(8,0).$
(c) $\pi'_{1}=\pi'_{2}=3,$ not zero; equal because $\hat\pi'=\pi'_{1}-\pi'_{2}=0$ there.
How did it go?
Stuck? Read 6. What a constraint does →
Mixed, as on the exam
Here the sections are mixed, and nothing says which method to use. Write the solution on paper, with reasons, before you look.
Let $f(x,y)=(x^{2}-1)(y-2).$
(a) Find all critical points of $f.$
(b) Show that none of them is a maximum or a minimum point, by looking at $f$ along two lines through each.
(c) Show that (10.2) fails at every point with $x\ne0.$
Hint
$f'_{2}=x^{2}-1,$ so $x=\pm1;$ then $f'_{1}=0$ fixes $y.$ For (b), write the point as $(1+s,\,2+t)$ and try $s=t$ and $s=-t.$
(a) The cases. $f'_{1}=2x(y-2)$ and $f'_{2}=x^{2}-1.$ From $f'_{2}=0,$ $x=1$ or $x=-1.$ Then $f'_{1}=\pm2(y-2)=0$ gives $y=2.$ The critical points are $(1,2)$ and $(-1,2),$ both with $f=0.$
(b) At $(1,2).$ With $x=1+s,$ $y=2+t{:}$ $f=(2s+s^{2})t.$ Along $s=t$ it is $2t^{2}+t^{3}=t^{2}(2+t)>0$ for $0<|t|<2;$ along $s=-t$ it is $-2t^{2}+t^{3}=t^{2}(t-2)<0$ for $0<|t|<2.$ Values above and below $0$ arbitrarily close: neither.
(b) At $(-1,2).$ With $x=-1+s,$ $y=2+t{:}$ $f=(s^{2}-2s)t.$ Along $s=t$ it is $t^{2}(t-2)<0,$ along $s=-t$ it is $t^{2}(t+2)>0,$ for $0<|t|<2.$ Neither again.
(c) The third condition. $f''_{11}=2(y-2),$ $f''_{22}=0,$ $f''_{12}=2x,$ so $$ \begin{aligned} &f''_{11}f''_{22}-\bigl(f''_{12}\bigr)^{2}\\ &=0-4x^{2}<0 \end{aligned} $$ for $x\ne0.$ So $f$ is neither concave nor convex on any convex set containing such a point; both critical points have $x=\pm1.$
Answer
(a) $(1,2)$ and $(-1,2).$
(b) At each, $f$ is positive along one line and negative along the other, arbitrarily close: neither.
(c) $f''_{11}f''_{22}-\bigl(f''_{12}\bigr)^{2}=-4x^{2}<0$ for $x\ne0.$
How did it go?
Stuck? Read 1. Where a maximum can be →
A firm's output is $F(K,L)=K^{1/4}L^{1/2}.$ It sells at $p=8,$ pays $r=1$ per unit of capital and $w=2$ per unit of labour, so for $K>0,$ $L>0$ $$\pi(K,L)=8K^{1/4}L^{1/2}-K-2L\,.$$
(a) Write down the conditions (10.1) and solve them.
(b) What is the marginal product of labour at the solution, and what does it equal in words?
(c) Show that the solution is the global maximum on $K>0,$ $L>0,$ and give the largest profit.
Hint
Divide one condition by the other to get $L$ in terms of $K.$ In (c), every second-order partial is a power of $K$ and $L$ times a constant: find the sign of each, for all $K,L>0.$
(a) The conditions. $pF'_{K}=r$ and $pF'_{L}=w$ read $$ \begin{aligned} 2K^{-3/4}L^{1/2}&=1\,,\\ 4K^{1/4}L^{-1/2}&=2\,. \end{aligned} $$
(a) Solve. The second gives $L^{1/2}=2K^{1/4}.$ In the first, $2K^{-3/4}\cdot2K^{1/4}=4K^{-1/2}=1,$ so $K=16,$ and $L^{1/2}=2\cdot16^{1/4}=4,$ so $L=16.$
(b) The real wage. $F'_{L}=\tfrac12K^{1/4}L^{-1/2} =\tfrac12\cdot2\cdot\tfrac14=\tfrac14,$ which is $w/p=2/8.$ The marginal product of labour equals the real wage.
(c) Concave on the quadrant. For all $K,L>0,$ $$ \begin{aligned} \pi''_{KK}&=-\tfrac32K^{-7/4}L^{1/2}<0\,,\\ \pi''_{LL}&=-2K^{1/4}L^{-3/2}<0\,,\\ \pi''_{KL}&=K^{-3/4}L^{-1/2}\,, \end{aligned} $$ and $$ \begin{aligned} &\pi''_{KK}\pi''_{LL}-\bigl(\pi''_{KL}\bigr)^{2}\\ &=3K^{-3/2}L^{-1}-K^{-3/2}L^{-1}\\ &=2K^{-3/2}L^{-1}>0\,. \end{aligned} $$
(c) Global. The quadrant $K>0,$ $L>0$ is a convex set, so $\pi$ is concave on it (Fact 10.11) and the interior critical point $(16,16)$ is a maximum point over the whole quadrant (Fact 10.13). There $F=2\cdot4=8,$ and $$ \begin{aligned} \pi(16,16)&=64-16-32\\ &=16\,. \end{aligned} $$
Answer
(a) $K=16,$ $L=16.$
(b) $F'_{L}=\tfrac14=w/p,$ the real wage.
(c) $\pi$ is concave on $K,L>0$ (the third condition is $2K^{-3/2}L^{-1}>0$), so $(16,16)$ is the global maximum; the largest profit is $16.$
How did it go?
Stuck? Read 5. Concavity decides, globally →
As on the exam
3 problems · 24 pProblems in exam form, with points. Solve them on paper or in the box. Then open the solution one step at a time and mark yourself against the rubric.
A quadratic: the global minimum, and the extremes on a segment
Let $$f(x,y)=x^{2}-xy+y^{2}-5x-2y\,.$$
- (a)3 p
Find all stationary points of $f$ and the value of $f$ at each.
- (b)3 p
Does $f$ have a global minimum on $\mathbb{R}^{2}$? A global maximum? Justify your answers.
- (c)4 p
Find the maximum and the minimum value of $f$ on the set $$ \begin{aligned} S=\{(x,y):\ &x+y=5,\\ &x\ge0,\ y\ge0\}\,, \end{aligned} $$ and the points where they are attained.
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(a) The first-order conditions. $$ \begin{aligned} f'_{1}&=2x-y-5=0\,,\\ f'_{2}&=-x+2y-2=0\,. \end{aligned} $$
(a) Solve. The first gives $y=2x-5;$ in the second, $-x+4x-10-2=0,$ so $3x=12,$ $x=4$ and $y=3.$ The only stationary point is $(4,3),$ with $$ \begin{aligned} f(4,3)&=16-12+9-20-6\\ &=-13\,. \end{aligned} $$
(b) The second-order partials are constants: $f''_{11}=2,$ $f''_{22}=2,$ $f''_{12}=-1.$ At every point of the plane $$ \begin{aligned} &f''_{11}=2\ge0\,,\\ &f''_{22}=2\ge0\,,\\ &f''_{11}f''_{22}-\bigl(f''_{12}\bigr)^{2}\\ &\quad=4-1=3\ge0\,. \end{aligned} $$
(b) A global minimum. So $f$ is convex on $\mathbb{R}^{2}$ (Fact 10.11), which is a convex set, and by Fact 10.13 the interior stationary point $(4,3)$ is a minimum point over the whole plane. The global minimum value is $f(4,3)=-13.$
(b) No global maximum. Along the $x$-axis $f(t,0)=t^{2}-5t,$ and for a natural number $n\ge6$ $$f(n,0)=n(n-5)\ge n\,.$$ So $f$ takes values larger than any number we name, and has no largest value.
(c) One variable on a closed interval. On $S$ we have $y=5-x$ with $0\le x\le5,$ and $$ \begin{aligned} g(x)&=f(x,5-x)\\ &=x^{2}-x(5-x)\\ &\quad+(5-x)^{2}\\ &\quad-5x-2(5-x)\\ &=3x^{2}-18x+15\,. \end{aligned} $$ $S$ is a segment (closed and bounded) and $g$ is continuous on $[0,5],$ so by the extreme value theorem (Fact 6.9) the maximum and the minimum exist.
(c) The candidates (Method 6.15): $g'(x)=6x-18=0$ at $x=3,$ inside $(0,5),$ and the two endpoints $x=0$ and $x=5{:}$ $$ \begin{aligned} g(3)&=27-54+15=-12\,,\\ g(0)&=15\,,\\ g(5)&=75-90+15=0\,. \end{aligned} $$ Check in $f{:}$ $f(3,2)=9-6+4-15-4=-12,$ $f(0,5)=25-10=15,$ $f(5,0)=25-25=0.$
(c) Compare. The maximum on $S$ is $15,$ at the endpoint $(0,5);$ the minimum on $S$ is $-12,$ at $(3,2).$ (It is larger than the global minimum $-13,$ as it must be: $(4,3)$ is not on $S.$)
Answer
(a) One stationary point, $(4,3),$ with $f(4,3)=-13.$
(b) $f$ is convex on $\mathbb{R}^{2},$ so the global minimum is $f(4,3)=-13.$ No global maximum ($f(n,0)=n(n-5)$ grows without bound).
(c) $\max_{S}f=15$ at $(0,5);$ $\min_{S}f=-12$ at $(3,2).$
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Tick each criterion that is really in your solution. A criterion scores only if what it asks for is there.
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Common errors
- The global minimum $-13$ reported as the minimum on $S.$ The point $(4,3)$ has $x+y=7,$ so it is not on the segment; on $S$ the function is $g(x)=3x^{2}-18x+15$ and its smallest value is $-12.$
- The endpoints left out in (c), so that no maximum is found, or the stationary point of $g$ called the maximum. $g''=6>0$ makes $x=3$ the minimum on the segment; the maximum is at an endpoint.
- “Global” claimed from the signs at $(4,3)$ alone. The verdict is global because the conditions of (10.2) hold at every point of the plane (Watch out 10.14).
- A sign slip with the cross term: $f'_{1}=2x+y-5$ instead of $2x-y-5.$ The system then gives $\bigl(\tfrac85,\tfrac95\bigr),$ which is not a stationary point of $f{:}$ there $f'_{1}=\tfrac{16}{5}-\tfrac95-5=-\tfrac{18}{5}.$ Differentiating the cross term $-xy$ once more, slowly, catches it.
Profit from two goods, with and without a capacity limit
A firm produces $x\ge0$ tonnes of one good and $y\ge0$ tonnes of another. Its profit, in thousands of kronor, is $$\pi(x,y)=17x+19y-x^{2}-xy-2y^{2}\,.$$
- (a)3 p
Find the plan $(x,y)$ that is the only candidate for maximum profit, and the profit there.
- (b)2 p
Show that this plan gives the largest profit of all plans.
- (c)3 p
Because of limited capacity the firm must produce exactly $6$ tonnes in total, so $x+y=6.$ Find the best plan and the largest profit under this restriction.
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(a) The first-order conditions. $$ \begin{aligned} \pi'_{1}&=17-2x-y=0\,,\\ \pi'_{2}&=19-x-4y=0\,. \end{aligned} $$
(a) Solve. The first gives $y=17-2x;$ in the second, $19-x-68+8x=0,$ so $7x=49,$ $x=7$ and $y=3.$ The only critical point is $(7,3),$ with $$ \begin{aligned} \pi(7,3)&=119+57\\ &\quad-49-21-18\\ &=88\,. \end{aligned} $$
(b) Concave everywhere. $\pi''_{11}=-2,$ $\pi''_{22}=-4,$ $\pi''_{12}=-1,$ constants. At every point $$ \begin{aligned} &\pi''_{11}=-2\le0\,,\\ &\pi''_{22}=-4\le0\,,\\ &\pi''_{11}\pi''_{22}-\bigl(\pi''_{12}\bigr)^{2}\\ &\quad=8-1=7\ge0\,. \end{aligned} $$
(b) Global. So $\pi$ is concave on the plane, a convex set (Fact 10.11), and by Fact 10.13 the critical point $(7,3)$ is a maximum point over the whole plane, so also over the plans with $x\ge0,$ $y\ge0.$ The largest profit is $88$ thousand kronor.
(c) Substitute. The best plan of (b) totals $10$ tonnes, so it is not available. On $x+y=6$ with $x\ge0,$ $y\ge0$ we have $y=6-x,$ $0\le x\le6,$ and $$ \begin{aligned} \hat\pi(x)&=17x+19(6-x)-x^{2}\\ &\quad-x(6-x)-2(6-x)^{2}\\ &=-2x^{2}+16x+42\,. \end{aligned} $$ It is continuous on $[0,6],$ so its largest value exists (Fact 6.9).
(c) The candidates. $\hat\pi'(x)=-4x+16=0$ at $x=4,$ inside $(0,6),$ and the endpoints: $$ \begin{aligned} \hat\pi(4)&=-32+64+42=74\,,\\ \hat\pi(0)&=42\,,\\ \hat\pi(6)&=-72+96+42=66\,. \end{aligned} $$
(c) Conclusion. The best plan is $x=4,$ $y=6-4=2,$ with profit $\pi(4,2)=68+38-16-8-8=74$ thousand kronor. (Scaling $(7,3)$ down to $6$ tonnes, $(4.2,1.8),$ gives $73.92{:}$ close, and not the best.)
Answer
(a) $(7,3),$ with profit $88$ thousand kronor.
(b) $\pi$ is concave on the whole plane ($-2\le0,$ $-4\le0,$ $8-1=7\ge0$ everywhere), so $(7,3)$ is the global maximum.
(c) $(4,2),$ with profit $74$ thousand kronor.
Mark your solution
Tick each criterion that is really in your solution. A criterion scores only if what it asks for is there.
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Common errors
- Under the capacity limit, the partials of $\pi$ set to zero and $x+y=6$ added: three equations in two unknowns, with no solution (Watch out 10.18). At the best plan $(4,2)$ the partials are $7$ and $7,$ not zero.
- The unconstrained plan scaled down to the new total: $(4.2,1.8)$ gives $73.92,$ less than $74.$ The best mix changes when the total does.
- In (b), “$\pi''_{11}<0$ and $\pi''_{22}<0,$ so it is a maximum”: the third condition is the one that decides (Example 10.7).
- The cross term dropped in (a): $\pi'_{1}=17-2x$ gives $x=8.5$ straight away, and the coupling between the two goods is lost.
All the stationary points, by cases
Find all stationary points of $$f(x,y)=xy\,(x+2y-8)$$ and the value of $f$ at each. (You do not need to decide what kind of points they are.)
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Expand first. $f(x,y)=x^{2}y+2xy^{2}-8xy.$
The partials, factored. $$ \begin{aligned} f'_{1}&=2xy+2y^{2}-8y\\ &=2y\,(x+y-4)\,,\\ f'_{2}&=x^{2}+4xy-8x\\ &=x\,(x+4y-8)\,. \end{aligned} $$
The cases. $f'_{1}=0$ when $y=0$ or $x+y=4.$ $f'_{2}=0$ when $x=0$ or $x+4y=8.$ Both vanish exactly when one condition of each pair holds: four cases.
Case $y=0.$ With $x=0{:}$ $(0,0).$ With $x+4y=8{:}$ $x=8,$ so $(8,0).$
Case $x+y=4.$ With $x=0{:}$ $y=4,$ so $(0,4).$ With $x+4y=8{:}$ subtracting $x+y=4$ gives $3y=4,$ so $y=\tfrac43$ and $x=4-\tfrac43=\tfrac83.$
The values. $f$ has the factor $xy,$ so $f(0,0)=f(8,0)=f(0,4)=0.$ And $$ \begin{aligned} f\bigl(\tfrac83,\tfrac43\bigr) &=\tfrac{32}{9}\bigl(\tfrac83+\tfrac83-8\bigr)\\ &=\tfrac{32}{9}\cdot\bigl(-\tfrac83\bigr)\\ &=-\tfrac{256}{27}\,. \end{aligned} $$
Answer
Four stationary points: $(0,0),$ $(8,0)$ and $(0,4),$ where $f=0,$ and $\bigl(\tfrac83,\tfrac43\bigr),$ where $f=-\tfrac{256}{27}.$
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Common errors
- Dividing $f'_{1}=0$ by $y,$ or $f'_{2}=0$ by $x,$ and so losing the points on the axes. A factor equal to zero is a case of its own.
- Combining the two factors of the same partial: $y=0$ with $x+y=4$ gives $(4,0),$ where $f'_{1}=0$ but $f'_{2}=-16\ne0.$ Each case takes one factor from $f'_{1}$ and one from $f'_{2}.$
- A slip in the product rule for $xy(x+2y-8)$; expanding first, $x^{2}y+2xy^{2}-8xy,$ avoids it.
- Classifying the points as maxima or minima by the second-order partials at each point: that test is Lecture 11's, and it is not asked for here.