16. Max-Min Problems

One first needs to recognize that, in this problem, the word "largest" means to maximize the volume of the box. Since all \(8\) vertices are on the ellipsoid, if one of the vertices is \((x,y,z)\) with \(x\), \(y\) and \(z\) positive, then the \(8\) are\((\pm x,\pm y,\pm z)\). So the length is \(L=2x\), the width is \(W=2y\), the height is \(H=2z\) and the volume is \[ V=LWH=(2x)(2y)(2z)=8xyz \] We need to maximize this volume while requiring the point \((x,y,z)\) to lie on the ellipsoid constraint: \[ x^2+\dfrac{y^2}{4}+\dfrac{z^2}{16}=1 \]

Now we need to solve the constraint for a variable and replace that variable in the optimal function by its value from the constraint. In this case, \(x\) is the easiest variable to replace, since it is easier to solve for: \[ x=\sqrt{1-\,\dfrac{y^2}{4}-\,\dfrac{z^2}{16}} \] Substituting into the volume gives \[ V=8yz\sqrt{1-\,\dfrac{y^2}{4}-\,\dfrac{z^2}{16}} \] This is the function we need to maximize. We set the partial derivatives equal to zero: \[\begin{aligned} V_y&=8z\sqrt{1-\,\dfrac{y^2}{4}-\,\dfrac{z^2}{16}} +\dfrac{8yz}{2\sqrt{1-\,\dfrac{y^2}{4}-\,\dfrac{z^2}{16}}} \left(\dfrac{-2y}{4}\right)=0 \qquad \text{(1)} \\ V_z&=8y\sqrt{1-\,\dfrac{y^2}{4}-\,\dfrac{z^2}{16}} +\dfrac{8yz}{2\sqrt{1-\,\dfrac{y^2}{4}-\,\dfrac{z^2}{16}}} \left(\dfrac{-2z}{16}\right)=0 \qquad \text{(2)} \end{aligned}\] and solve for \(x\) and \(y\). We start by clearing the denominators in equations (1) and (2): \[\begin{aligned} &16z\left(1-\,\dfrac{y^2}{4}-\,\dfrac{z^2}{16}\right)-4y^2z=0 \\ &16y\left(1-\,\dfrac{y^2}{4}-\,\dfrac{z^2}{16}\right)-yz^2=0 \end{aligned}\] and simplify: \[\begin{aligned} &16z-4y^2z-z^3-4y^2z=0 \\ &16y-4y^3-yz^2-yz^2=0 \end{aligned}\] or: \[\begin{aligned} &z(16-8y^2-z^2)=0 \\ &2y(8-2y^2-z^2)=0 \end{aligned}\] Notice that \(y\) and \(z\) cannot be \(0\) or else the volume would be \(0\). So the equations say: \[\begin{aligned} &8y^2+z^2=16 \qquad \text{(3)} \\ &2y^2+z^2=8 \qquad \enspace \text{(4)} \end{aligned}\] Equation (3) minus equation (4) says \(6y^2=8\). So \(y=\dfrac{2}{\sqrt{3}}\). Similarly, \(4\) times equation (4) minus equation(3) says \(3z^2=16\). So \(z=\dfrac{4}{\sqrt{3}}\). Consequently, \[ x=\sqrt{1-\,\dfrac{y^2}{4}-\,\dfrac{z^2}{16}} =\sqrt{1-\,\dfrac{1}{3}-\,\dfrac{1}{3}}=\dfrac{1}{\sqrt{3}} \] Finally, we carefully reread the problem. We need to find the dimensions and the volume. The length, width and height are: \[ L=2x=\dfrac{2}{\sqrt{3}} \qquad W=2y=\dfrac{4}{\sqrt{3}} \qquad H=2z=\dfrac{8}{\sqrt{3}} \] So the volume is: \[ V=LWH=\dfrac{64}{3\sqrt{3}} \]

It is important to recognize that the word "closest" means that we should minimize the distance. In this case, we need to minimize the distance from the point \(P=(7,1,4)\) to a general point \(Q=(x,y,z)\) on the plane. This distance is: \[ D=\sqrt{(x-7)^2+(y-1)^2+(z-4)^2} \] Now if the distance is a minimum, then so is the square of the distance. So we will actually minimize the function: \[ f=D^2=(x-7)^2+(y-1)^2+(z-4)^2 \]

The constraint is the equation of the plane, which we solve for \(z\): \[ z=2-3x+y \] Substituting into the extremal, we need to minimize \[ f=(x-7)^2+(y-1)^2+(-3x+y-2)^2 \] We set the partials equal to \(0\): \[\begin{aligned} f_x&=2(x-7)+2(-3x+y-2)(-3)=0 \\ f_y&=2(y-1)+2(-3x+y-2)=0 \end{aligned}\] We divide these equations by \(2\) and simplify: \[\begin{aligned} 10x-3y&=1 \\ -3x+2y&=3 \end{aligned}\] We solve these equations to find \(x=1\) and \(y=3\) and so \(z=2-3x+y=2-3(1)+(3)=2\). So the point is: \[ (x,y,z)=(1,3,2) \] If we were asked for the distance, it would be: \[\begin{aligned} D&=\sqrt{(1-7)^2+(3-1)^2+(2-4)^2} \\ &=\sqrt{(6)^2+(2)^2+(2)^2}=\sqrt{44} \end{aligned}\] Don't forget the square root!