Calculus Maximum or Minimum: cylindrical boiler, flat base, hemispherical top; length h meters, top's radius r meters. Find ratio of r to h so that...

[reggiwilliams]

Can you please look at question and my solution
Think it’s right
I have assumed area includes area of base.
Not great with problems of this nature and seems to take forever at getting to answer.
Will practice doing similar questions.

 

[blamocur]

I do get the same answer, i.e. [imath]r=h[/imath], but I used a different approach:
Both [imath]V[/imath] and [imath]A[/imath] are functions of two variables. For [imath]V[/imath] to have an extremum point on the contour [imath]A(r,h)=\text{const}[/imath] we need the gradient of [imath]V[/imath] to be orthogonal to that contour, which means that the gradients of [imath]V[/imath] and [imath]A[/imath] must be aligned, i.e.
[math]\frac{\partial A}{\partial r} \frac{\partial V}{\partial h} = \frac{\partial A}{\partial h}\frac{\partial V}{\partial r}[/math]While this might look intimidating the resulting equation is actually simpler. Note this is very similar to using Lagrange multipliers.

 

[blamocur]

Can you please look at question and my solution
Think it’s right
I have assumed area includes area of base.
Not great with problems of this nature and seems to take forever at getting to answer.
Will practice doing similar questions.

Another minor simplification: since every expression has [imath]\pi[/imath] in it you can factor it out, i.e., use constant [imath]\pi k[/imath] instead of [imath]k[/imath], and look for the maximum of [imath]\frac{V}{\pi}[/imath] instead of [imath]V[/imath]. The same is applicable to my approach in post #2. In both approaches it makes the intermediate expressions a little bit less hairy.