EXTREME VALUES AND OPTIMIZATION

Question

Solve the given optimization problem below. Show your complete solution.

Find the radius of a closed cylindrical tank having a maximum volume if the material available for its construction is 45 in^2.

Answer 1

Given:

S.A. = 45 in²

Find:

r = ? in to maximize V

Solution:

SA = 2πr² + 2πrh

V(r, h) = πr²h

But SA is a constant, limiting value, thus, h is going to be a function of r:

2πrh = SA - 2πr²

h( r ) = SA / (2πr) - r

V(r) = πr² ( SA / (2πr) - r )

V(r) = r / 2 * SA - πr³

V'(r) = SA / 2 - 3πr² = 0

SA - 6πr² = 0

SA = 6πr²

r² = SA / 6π

∴ r = √( SA / 6π ) ≈ 1.545 in

Note: I omit the ± because negative radius would not make sense in this particular scenario since we're not in polar coordinate systems or anything where negative radii could matter.

I hope this helps! Message me in the comments with any questions, comments, or concerns!

To ensure true maximum, check V''( r = 1.545 ) to be negative, however, V''(r) = -6πr, which means that if I had included the -√( SA / 6π ), it would be a minimum.

V''(r) < 0 means maximum

V''(r) > 0 means minimum