Best Answer:
The dimensions we need to worry about here is the diameter of the base and the height.

This value cannot be more than 158 cm. Let's set up this equation:

158 = d + h

In terms of radius:

158 = 2r + h

Let's solve this for r in terms of h:

158 - h = 2r

79 - h/2 = r

We want a maximum volume of the cylinder, so let's get that equation:

V = πr²h

If we substitute the expression above into this equation we will get volume in terms of only the height:

V = π(79 - h/2)²h

Simplify:

V = π(6241 - 79h + h²/4)h

V = 6241πh - 79πh² + (π/4)h³

The maximum volume happens when the first derivative is zero, so:

dV/dh = 6241π - 158πh + (3π/4)h²

0 = 6241π - 158πh + (3π/4)h²

Divide both sides by π, then put into standard polynomial form:

0 = 6241 - 158h + (3/4)h²

0 = (3/4)h² - 158h + 6241

Let's simplify further by multiplying both sides by 4:

0 = 3h² - 632h + 24964

Now we can solve for this quadratic and throw out any values that don't make sense:

h = [ -b ± √(b² - 4ac)] / (2a)

h = [ -(-632) ± √((-632)² - 4(3)(24964))] / (2 * 3)

h = [ 632 ± √(399424 - 299568)] / 6

h = [ 632 ± √(99856)] / 6

h = (632 ± 316) / 6

h = 316 / 6 and 948 / 6

h = 158/3 and 158

Since a height of 158 would result in a radius of 0, a volume of 0, this is a minimum so we'll throw that out.

h = 158/3

Now that we have h, solve for r:

r = 79 - h/2

r = 79 - (158/3)/2

r = 79 - 158/6

r = 474/6 - 158/6

r = 316/6

r = 158/3

Solving for d now:

d = 2r

d = 2(158/3)

d = 316/3

The maximum volume would have a diameter of 316/3 (105.33) cm and a height of 158/3 (52.67) cm

The sum of this diameter and height will be 158 cm:

105.33 + 52.67 = 158

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