0

I'm stuck... The problem states that - You wish to construct a right cylindrical can (with bottom and top) from a square steel plate with side length 10 cm. What dimensions on r and h maximizes the volume of the can.

Can someone please point me in the right direction.

mr.hyde
  • 1

1 Answers1

1

Use the fact that the formula for the surface area of a right circular cylinder is $A=2\pi r h + 2 \pi r^2$:

$$ 10^2=2\pi r h + 2 \pi r^2 $$

Then express $h$ as a function of $r$ and use the formula for the volume of a right circular cylinder: $$ h(r)=\frac{50}{\pi r}-r\\ V(r)=\pi r^2 h(r) = \pi r^2 \left(\frac{50}{\pi r}-r \right)=50r-\pi r^3,\ 0\lt r\lt \sqrt\frac{50}{\pi}$$

Now, just find what $r$ makes the function $V(r)$ attain the maximum value. For that, set the first derivative of $V(r)$ equal to zero and solve for $r$: $V'(r)=50-3\pi r^2=0\implies r=\sqrt{\frac{50}{3\pi}}$. When you find the $r$, it's going to be trivial to find $h$: $h\left(\sqrt{\frac{50}{3\pi}}\right)$.

Michael Rybkin
  • 6,646
  • 2
  • 11
  • 26
  • When i do this, i get the sqare root of A/6pi for r. Which is wrong apparently... – mr.hyde Apr 05 '19 at 13:53
  • 1
    In fact, I think that the problem is much more difficult that it looks. $100$ is not the surface area of the can but the area of the plate ! Interesting problem. – Claude Leibovici Apr 05 '19 at 14:03
  • Well, you have a square steel plate whose side is $10$ cm. Its area is going to be 100 square cementers. That's the amount of material that you're allowed to use to construct the surface area of the can. Am I not right? – Michael Rybkin Apr 05 '19 at 14:07
  • You are correct, theres going to be excess steel. – mr.hyde Apr 05 '19 at 14:24
  • according to the answer in the book the r should be 5/π and h = 10(1-1/π). And i´m either retarded or blind. Because i can´t seem to get the right answer – mr.hyde Apr 05 '19 at 14:54
  • Hmm, I do think I have set the problem up correctly. – Michael Rybkin Apr 05 '19 at 14:59