   Chapter 14.7, Problem 55E

Chapter
Section
Textbook Problem

If the length of the diagonal of a rectangular box must be L, what is the largest possible volume?

To determine

To find: The largest volume of the rectangular box.

Explanation

Given:

The length of the diagonal of the rectangular box is L .

Calculation:

Let the dimensions be x,yandz .

Then, the volume of the rectangular box is V(x,y,z)=xyz where x>0,y>0,z>0 .

The length of the diagonal of the rectangular box is L=x2+y2+z2 .

Thus, the maximum value of the function V(x,y,z)=xyz subject to the equation L=x2+y2+z2 .

The value of z is calculated from the equation L=x2+y2+z2 .

L2=x2+y2+z2z2=L2x2y2z=L2x2y2

Substitute z=L2x2y2 in the function V=xyz .

V=xy(L2x2y2)

Take the partial derivative of the function V with respect x and obtain Vx .

Vx=x(xy(L2x2y2))=yx(xL2x2y2)=y[L2x2y2(1)+x(12)(L2x2y2)12(2x)]=yL2x2y2x2yL2x2y2

Hence, Vx=yL2x2y2x2yL2x2y2 . (1)

Take the partial derivative of the function V with respect y and obtain Vy .

Vy=y(xy(L2x2y2))=xy(yL2x2y2)=x[L2x2y2(1)+y(12)(L2x2y2)12(2y)]=xL2x2y2xy2L2x2y2

Thus, Vy=xL2x2y2xy2L2x2y2 . (2)

Set the above partial derivatives to 0 and find the critical points.

From the equation (1),

yL2x2y2x2yL2x2y2=0y(L2x2y2)x2yL2x2y2=0y(L2x2y2)x2y=0y(L2x2y2x2)=0

Simplify further as follows

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