A diagram of to model this problem is shown below
The volume of a cone is given
V= 1/3(πr²h)
We will use some Calculus to find the maximum volume of the cone, but first, we need to get V in one term only, either in term of 'r' or in term of 'h'
Referring to the diagram, we can start with the right-angle triangle that has two short sides; the radius of the cone and the (h-r) and the hypotenuse 'R' which is the radius of the sphere.
Forming an equation based on Pythagoras theorem
r² = R² - (h-R)² ⇒ We know that R = 8
r² = 8² - (h - 8)²
r² = 64 - (h² - 16h + 64)
r² = 64 - h² + 16h - 64
r² = 16h - h²
Substitute r² = 16h - h² into the formula for V, we have
V = 1/3 (π) (16h - h²) h
V = 1/3 (π) (16h² - h³)
V = (16/3)πh² - (1/3)πh³
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Now we have V in term of 'h', we will use Calculus to find the value of 'h' that will give the maximum value of V
First, we will differentiate V, then set dV/dh = 0
V = (16/3)πh² - (1/3)πh³
dV/dh = (32/3)πh - πh²
Setting dV/dh = 0
0 = (32/3)πh - πh²
0 = πh [(32/3) - h]
πh = 0 and (32/3) - h = 0
h = 0 and h = 32/3
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The value of h that we will choose is h = 32/3 since h = 0 is unlikely
Substituting h = 32/3 back into V, we have
V = 1/3 (π) [16 (32/3)² - (32/3)³]
V = 1/3 π × 606.814
V = 635.5 units³
