The values of h and r to maximize the volume are r = 4 and h = 2
<h3>The formula for h in terms of r</h3>
From the question, we have the following equation
2r + 2h = 12
Divide through by 2
r + h = 6
Subtract r from both sides of the equation
h = 6 - r
Hence, the formula for h in terms of r is h = 6 - r
<h3>Formulate a function V(r)</h3>
The volume of a cylinder is
V = πr²h
Substitute h = 6 - r in the above equation
V = πr²(6 - r)
Hence, the function V(r) is V = πr²(6 - r)
<h3>The single critical point</h3>
V = πr²(6 - r)
Expand
V = 6πr² - πr³
Integrate
V' = 12πr - 3πr²
Set to 0
12πr - 3πr² = 0
Divide through by 3π
4r - r² = 0
Factor out r
r(4 - r) = 0
Divide through by 4
4 - r = 0
Solve for r
r = 4
Hence, the single critical point on the interval [0. 6] is r = 4
<h3>Prove that the critical point is a global maximum</h3>
We have:
V = πr²(6 - r)
and
V' = 12πr - 3πr²
Determine the second derivative
V'' = 12π - 6πr
Set r = 4
V'' = 12π - 6π* 4
Evaluate the product
V'' = 12π - 24π
Evaluate the difference
V'' = -12π
Because V'' is negative, then the single critical point is a global maximum
<h3>The values of h and r to maximize the volume</h3>
We have
r = 4 and h = 6 - r
Substitute r = 4 in h = 6 - r
h = 6 - 4
Evaluate
h = 2
Hence, the values of h and r to maximize the volume are r = 4 and h = 2
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