16. Max-Min Problems

Homework

Consider the function $f(x,y)=x^3-xy-x+y^2$. Find all critical points. Then use the Second Derivative Test to classify each as a local minimum, local maximum or saddle or say the test fails.

We compute the partial derivatives and set them equal to $0$: \[\begin{aligned} f_x&=3x^2-y-1=0 \\ f_y&=-x+2y=0 \end{aligned}\] We solve for $x$ and $y$: Since $x=2y$, we have $12y^2-y-1=0$. We use the quadratic formula: \[ y=\dfrac{1\pm\sqrt{1+48}}{24}=\dfrac{1\pm7}{24}=\dfrac{1}{3},\dfrac{-1}{4} \] If $y=\dfrac{1}{3}$ then $x=\dfrac{2}{3}$. If $y=\dfrac{-1}{4}$ then $x=\dfrac{-1}{2}$.
The second derivatives and the discriminant are: \[ f_{xx}=6x, \quad f_{yy}=2, \quad f_{xy}=-1, \quad D=f_{xx}f_{yy}-{f_{xy}}^2=12x-1 \] At $(x,y)=\left(\dfrac{2}{3},\dfrac{1}{3}\right)$: \[ f_{xx}=4, \quad D=7, \qquad \text{which is a local minimum.} \] At $(x,y)=\left(\dfrac{-1}{2},\dfrac{-1}{4}\right)$: \[ f_{xx}=-3, \quad D=-7, \qquad \text{which is a saddle point.} \]
Find the radius, height and volume of the largest cylinder with base on the $xy$ plane and upper circle on the cone $z=3-\sqrt{x^2+y^2}$.
Work in cylindrical coordinates.
1. Solve by eliminating $z$.
  
  In cylindrical coordinates, the cone is $z=3-r$. So the volume of the cylinder is \[ V=\pi r^2h=\pi r^2z=\pi r^2(3-r)=\pi(3r^2-r^3) \] To maximize $V$, we set the derivative equal to $0$ and solve for $r$: \[ V'=\pi(6r-3r^2)=3\pi r(2-r)=0 \quad \implies \quad r=0,2 \] $r=0$ gives $V=0$. So the only critical point with non-zero volume is $r=2$ which must produce the maximum volume. Then $z=3-r=3-2=1$ and $V=\pi r^2z=\pi 2^2 1=4\pi$.
2. Solve by Lagrange multipliers.
  
  In cylindrical coordinates, the cone is $z=3-r$. The extremal is the volume of the cylinder: \[ V=\pi r^2h=\pi r^2z \] The constraint is the equation of the cone, written as: \[ g=z+r=3 \] Taking the variables in the order $r,z$, the gradients are: \[ \vec\nabla V=\langle2\pi rz,\pi r^2\rangle \quad \text{and} \quad \vec\nabla g=\langle1,1\rangle \] The Lagrange equations $\vec\nabla V=\lambda\vec\nabla g$ are: \[ 2\pi rz=\lambda \quad \text{and} \quad \pi r^2=\lambda \] Eliminating $\lambda$ gives $2\pi rz=\pi r^2$. Since $r\ne0$, this says $z=\dfrac{r}{2}$. Plugging this into the constraint gives: \[ 3=z+r=\dfrac{r}{2}+r=\dfrac{3}{2}r \] Therefore, $r=2$, $z=1$ and $V=\pi r^2z=4\pi$.
Find the dimensions and volume of the largest rectangular box with base on the $xy$ plane and upper vertices on the cone $z=3-\sqrt{x^2+y^2}$.
Work in rectangular coordinates.

$The figure shows a rectangular box sitting on the xy plane with its upper vertices on cone z=3-\sqrt{x^2+y^2}. The box is animated so that its length and width oscillate. When the length or width become 0, the volume is 0. Somewhere in between these extremes is a rectangular box with maximal volume.$
1. Solve by Lagrange multipliers.
  HINT: Rewrite the constraint to eliminate the square root.
  
  We can rewite the equation of the cone as $\sqrt{x^2+y^2}=3-z$ or $x^2+y^2=(z-3)^2$. So we take the constraint as \[ g=x^2+y^2-(z-3)^2=0 \] We take $(x,y)$ in the first quadrant. Then the volume of the box is \[ V=LWH=(2x)(2y)(z)=4xyz \] The gradients are: \[ \vec\nabla V=\langle4yz,4xz,4xy\rangle \quad \text{and} \quad \vec\nabla g=\langle2x,2y,-2(z-3)\rangle \] The Lagrange equations $\vec\nabla V=\lambda\vec\nabla g$ are: \[ 4yz=\lambda2x \qquad 4xz=\lambda2y \qquad 4xy=-\lambda2(z-3) \] The left sides become the same if we multiply the first by $x$, the second by $y$ and the third by $z$. \[ 4xyz=\lambda2x^2=\lambda2y^2=-\lambda2z(z-3) \] These say: \[ x^2=y^2=-z(z-3) \] Plugging this into the constraint gives: \[\begin{aligned} x^2+y^2-(z-3)^2&=0 \\ -2z(z-3)-(z-3)^2&=0 \\ -(z-3)(2z+z-3)&=0 \\ -3(z-3)(z-1)&=0 \\ \end{aligned}\] So $z=1$ or $z=3$. If $z=3$, then $x=y=0$ and so $V=0$. So the only critical point with non-zero volume is $z=1$ which must produce the maximum volume. Then $x^2=y^2=2$ and $x=y=\sqrt{2}$ and $V=4xyz=4(\sqrt{2})^2(1)=8$.
2. Solve by eliminating $z$.
  HINT: What is $xV_x-yV_y$?
  
  We take $(x,y)$ in the first quadrant. Then the volume of the box is \[ V=LWH=(2x)(2y)(z)=4xyz \] We substitute the equation of the cone $z=3-\sqrt{x^2+y^2}$: \[ V=4xy(3-\sqrt{x^2+y^2})=12xy-4xy\sqrt{x^2+y^2} \] We set the partial derivatives equal to $0$ and solve: \[\begin{aligned} V_x&=12y-4y\sqrt{x^2+y^2}-\dfrac{4x^2y}{\sqrt{x^2+y^2}}=0 \\ V_y&=12x-4x\sqrt{x^2+y^2}-\dfrac{4xy^2}{\sqrt{x^2+y^2}}=0 \\ \end{aligned}\] We multiply the first equation by $x$ and the second by $y$ and subtract: \[ \dfrac{4x^3y-4xy^3}{\sqrt{x^2+y^2}}=0 \] Since $x \gt 0$ and $y \gt 0$, this says $x=y$. Then the $V_x$ equation reduces to \[\begin{aligned} 12x-4x\sqrt{2x^2}-\dfrac{4x^3}{\sqrt{2x^2}}&=0 \\ 12x-4\sqrt{2}x^2-2\sqrt{2}x^2&=0 \\ 12x-6\sqrt{2}x^2&=0 \\ 2x-\sqrt{2}x^2&=0 \\ \sqrt{2}x(\sqrt{2}-x)&=0 \end{aligned}\] Then $x=0$ which gives $V=0$. Or $x=\sqrt{2}$. Then $y=\sqrt{2}$ and $z=3-\sqrt{x^2+y^2}=3-\sqrt{2+2}=1$ and $V=4xyz=4(\sqrt{2})^2(1)=8$.
Find the point that is closest to the origin on the surface $z=\dfrac{108}{x^4y^6}$ in the first octant. Then find the distance from the point to the origin.

We need to minimize the distance from the origin to the point $(x,y,z)$ on the surface: \[ D=\sqrt{x^2+y^2+z^2} \] It is easier to minimize the square of the distance, which will give the same solution. So we will minimize the function \[ f=D^2=x^2+y^2+z^2 \] We write the surface as the constraint: \[ g=x^4y^6z=108 \] The gradients are: \[ \vec\nabla f=\langle2x,2y,2z\rangle \quad \text{and} \quad \vec\nabla g=\langle 4x^3y^6z,6x^4y^5z,x^4y^6 \rangle \] The Lagrange equations $\vec\nabla f=\lambda\vec\nabla g$ are: \[ 2x=\lambda 4x^3y^6z \qquad 2y=\lambda 6x^4y^5z \qquad 2z=\lambda x^4y^6 \] The right sides become the same if we multiply the first by $6x$, the second by $4y$ and the third by $24z$: \[ \lambda24x^4y^6z=12x^2=8y^2=24z^2 \] This says $x^2=2z^2$ and $y^2=3z^2$. We substitute this into the constraint: \[ 108=x^4y^6z=(x^2)^2(y^2)^3z=(2z^2)^2(3z^2)^3z=108z^{11} \] So $z=1$ and consequently $x=\sqrt{2}$ and $y=\sqrt{3}$. So the closest point is $(x,y,z)=(\sqrt{2},\sqrt{3},1)$. Its distance from the origin is \[ D=\sqrt{2+3+1}=\sqrt{6} \]
A $118\,\text{cm}$ wire is cut into $3$ pieces of lengths $a$, $b$ and $c$. The piece of length $a$ is bent into a square. The piece of length $b$ is bent into a rectangle whose length is twice its width. The piece of length $c$ is bent into a rectangle whose length is $4$ times its width. What are $a$, $b$ and $c$ which minimize the total area enclosed by all $3$ pieces?
Solve by using Lagrange Multipliers.
$50\%$ off for solving by Eliminating a Variable.
$50\%$ Extra Credit for solving both ways. Be sure to clearly separate the calculations.

We express the length of the wire as the sum of its parts: \[ L(a,b,c) = a + b +c = 118 \] To minimize the total area of the shapes, we find the area of each shape as a function of the wire segment lengths.
The piece of length $a$ is folded into a square of side $s$: \[\begin{aligned} \text{Perimeter}_{a} &= a = 4s \\ \text{Area}_{a} &= s^2= \dfrac{a^2}{16} \end{aligned}\] The piece of length $b$ is folded into a rectangle of width $w$ and length $2w$: \[\begin{aligned} \text{Perimeter}_{b} &= b = 2w + w + 2w + w = 6w \\ \text{Area}_{b} &= 2w \cdot w = 2w^2= \dfrac{b^2}{18} \end{aligned}\] The piece of length $c$ is folded into a rectangle of width $W$ and length $4W$: \[\begin{aligned} \text{Perimeter}_{c} &= c = 4W + W +4W + W = 10W \\ \text{Area}_{c} &= 4W \cdot W = 4W^2= \dfrac{c^2}{25} \end{aligned}\] The total area is then found by adding the three areas: \[\begin{aligned} A &= \dfrac{a^2}{16} + \dfrac{b^2}{18}+ \dfrac{c^2}{25}\\ \end{aligned}\]

Lagrange Multipliers: The gradients of the functions are \[ \vec\nabla A=\left(\dfrac{a}{8}, \dfrac{b}{9}, \dfrac{2c}{25}\right) \qquad \vec\nabla L=\left(1,1,1\right) \] The Lagrange equations, $\vec\nabla A=\lambda\vec\nabla L$, are \[ \dfrac{a}{8}=\lambda \qquad \dfrac{b}{9}=\lambda \qquad \dfrac{2c}{25}=\lambda \] Solving for $b$ and $c$ in terms of $a$ gives: \[\begin{aligned} \dfrac{b}{9}&=\dfrac{a}{8} &\implies b=\dfrac{9a}{8} \\ \dfrac{2c}{25}&=\dfrac{a}{8} &\implies c=\dfrac{25a}{16} \end{aligned}\] Plugging these results into the constraint $L=118$ gives: \[\begin{aligned} a + \dfrac{9a}{8} + \dfrac{25a}{16} &= 118 \\ 16a+18a+25a&= 16\cdot 118 \\ 59a&= 16\cdot 118 \\ a &= 32 \end{aligned}\] And using this value in previous equations yields: \[ b = 36 \qquad c = 50 \] So the critical point is $(a,b,c)=(32,36,50)$ and the area is: \[\begin{aligned} A&=\dfrac{a^2}{16} + \dfrac{b^2}{18} + \dfrac{c^2}{25} \\ &=\dfrac{32^2}{16} + \dfrac{36^2}{18} + \dfrac{50^2}{25} \\ &=64+72+100=236 \end{aligned}\]

Eliminating a Variable:
We solve the constraint for $c=118-a-b$ and plug into the area: \[ A=\dfrac{a^2}{16} + \dfrac{b^2}{18}+ \dfrac{(118-a-b)^2}{25} \] The partial derivatives are: \[\begin{aligned} A_a&=\dfrac{a}{8} - \dfrac{2}{25}(118-a-b)=0 \\ A_b&=\dfrac{b}{9} - \dfrac{2}{25}(118-a-b)=0 \end{aligned}\] Multiply the first by $8\cdot25$ and the second by $9\cdot25$ and solve: \[\begin{aligned} 25a - 16(118-a-b)&=0 \\ 25b - 18(118-a-b)&=0 \end{aligned}\] \[\begin{aligned} 41a + 16b &= 16(118) \\ 18a + 43b &= 18(118) \end{aligned}\] \[\begin{aligned} 43\cdot41a + 43\cdot16b &= 43\cdot16(118) \\ 16\cdot18a + 16\cdot43b &= 16\cdot18(118) \end{aligned}\] \[\begin{aligned} (43\cdot41-16\cdot18)a&= (43-18)\cdot16(118) \\ 1475a&= 25\cdot16(118) \\ 59a&=16(118) \\ a&=32 \\ \end{aligned}\] We substitute back: \[\begin{aligned} 16b&=16(118)-41a=16(118)-41(32) \\ b&=118-82=36 \\ c&=118-a-b=118-32-36=50 \end{aligned}\] So the minimum area is $A=236$.

The maximum occurs when all the area is in one shape.
If $a=b=0$, then $c=118$ and $A=\dfrac{c^2}{25}=\dfrac{118^2}{25}\approx557$.
If $a=c=0$, then $b=118$ and $A=\dfrac{b^2}{18}=\dfrac{118^2}{18}\approx774$.
If $b=c=0$, then $a=118$ and $A=\dfrac{a^2}{16}=\dfrac{118^2}{16}\approx870$.
So the maximum occurs when there is only the square.
A box with no lid is to hold $108\,\text{cm}^3$. Its bottom costs $\text{\textdollar} 20$ per $\text{cm}^2$. Its left and right sides cost $\text{\textdollar} 10$ per $\text{cm}^2$. Its front and back cost $\text{\textdollar} 5$ per $\text{cm}^2$. Find the dimensions of the box with minimal cost.

Let $x$ be the length from left to right. Let $y$ be the width from front to back. Let $z$ be the height. So the cost to be minimized is \[\begin{aligned} C&=\text{\textdollar}20xy+2\cdot\text{\textdollar}10yz+2\cdot\text{\textdollar}5xz \\ &=\text{\textdollar}20xy+\text{\textdollar}20yz+\text{\textdollar}10xz \end{aligned}\] The constraint is the volume \[ V=xyz=108 \] The gradients of the functions are \[\begin{aligned} \vec\nabla C&=\langle 20y+10z, 20x+20z, 20y+10x \rangle \\ \vec\nabla V&=\langle yz, xz, xy \rangle \end{aligned}\] The Lagrange equations, $\vec\nabla C=\lambda\vec\nabla V$, are \[\begin{aligned} 20y+10z&=\lambda yz \qquad &\text{(1)} \\ 20x+20z&=\lambda xz \qquad &\text{(2)} \\ 20y+10x&=\lambda xy \qquad &\text{(3)} \end{aligned}\] The right sides become the same if we multiply (1)) by $x$, (2) by $y$ and (3) by $z$: \[ \lambda xyz=20xy+10xz=20xy+20yz=20yz+10xz \] The first equality says $10xz=20yz$ or $x=2y$. The second equality says $20xy=10xz$ or $z=2y$. Plugging these into the constraint gives: \[ V=xyz=(2y)y(2y)=108 \quad \text{or} \quad y^3=27 \quad \text{or} \quad y=3\,\text{cm} \] Then $x=z=6\,\text{cm}$. So the minimum cost is \[\begin{aligned} C&=20xy+20yz+10xz \\ &=20(6)(3)+20(3)(6)+10(6)(6) \\ &=360+360+360 =\text{\textdollar}1080 \end{aligned}\]
A quartic ellipsoid is the graph of the equation: \[ \dfrac{x^4}{a^4}+\dfrac{y^4}{b^4}+\dfrac{z^4}{c^4}=1 \] where $a$. $b$ and $c$ are the intercepts.
Find the dimensions and volume of the largest rectangular solid that can fit inside the quartic ellipsoid: \[ 16x^4+81y^4+z^4=3 \]

Let $(x,y,z)$ be the corner of the box in the first octant. Then the volume is \[ V=LWH=(2x)(2y)(2z)=8xyz \] The constraint is the equation of the surface: \[ g=16x^4+81y^4+z^4=3 \] The gradients of the functions are \[\begin{aligned} \vec\nabla V&=\langle 8yz, 8xz, 8xy \rangle \\ \vec\nabla g&=\langle 64x^3, 324y^3, 4z^3 \rangle \end{aligned}\] The Lagrange equations, $\vec\nabla V=\lambda\vec\nabla g$, are \[\begin{aligned} 8yz&=\lambda 64x^3 \qquad &\text{(1)} \\ 8xz&=\lambda 324y^3 \qquad &\text{(2)} \\ 8xy&=\lambda 4z^3 \qquad &\text{(3)} \end{aligned}\] The left sides become the same if we multiply (1) by $x$, (2) by $y$ and (3) by $z$: \[ 8xyz=\lambda 64x^4=\lambda 324y^4=\lambda 4z^4 \] The three right sides give $x$ and $y$ in terms of $z$: \[\begin{aligned} 64x^4&=4z^4 &\implies 16x^4&=z^4 &\implies x&=\dfrac{z}{2} \\ 324y^4&=4z^4 &\implies 81y^4&=z^4 &\implies y&=\dfrac{z}{3} \\ \end{aligned}\] Substituting this into the volume constraint gives \[ 3=16x^4+81y^4+z^4=z^4+z^4+z^4=3z^4 \] Therefore, $z=1$, $x=\dfrac{1}{2}$ and $ y=\dfrac{1}{3}$. So the dimensions and volume are: \[ L=2x=1, \qquad W=2y=\dfrac{1}{3}, \qquad H=2z=2, \qquad V=LWH=\dfrac{2}{3} \]

16. Max-Min Problems

Homework

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