   Chapter 14.7, Problem 47E

Chapter
Section
Textbook Problem

Find the maximum volume of a rectangular box that is inscribed in a sphere of radius r.

To determine

To find: The maximum volume of a rectangular box that is inscribed in a sphere.

Explanation

Let the equation of a sphere x2+y2+z2=r2 where x, y, z are parameters and r is the radius of the sphere.

Then the volume of the rectangular box is V=8xyz where the dimensions are 2x, 2y and 2z.

Thus, the maximum value of the function V(x,y,z)=8xyz subject to the equation x2+y2+z2=r2 .

The value of z is calculated from the equation, x2+y2+z2=r2 .

z2=r2x2y2z=r2x2y2

Substitute the value of z=r2x2y2 in the function V=8xyz ,

V=8xy(r2x2y2) .

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

Vx=8x(xy(r2x2y2))=8yx(xr2x2y2)=8y[r2x2y2(1)+x(12)(r2x2y2)12(2x)]=8y[r2x2y2x2r2x2y2]

Simplify further as follows.

Vx=8y[r2x2y2x2r2x2y2]=8y[r2x2y2x2r2x2y2]=8y[r22x2y2r2x2y2]=8y(r22x2y2)r2x2y2

Hence, Vx=8y(r22x2y2)r2x2y2 (1)

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

Vy=y(8xy(r2x2y2))=8xy(yr2x2y2)=8x[r2x2y2(1)+y(12)(r2x2y2)12(2y)]=8x[r2x2y2y2r2x2y2]

Simplify further as follows.

Vy=8x[r2x2y2y2r2x2y2]=8x[r2x2y2y2r2x2y2]=8x[r22y2x2r2x2y2]=8x(r22y2x2)r2x2y2

Thus, Vy=8x(r22y2x2)r2x2y2 (2)

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

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