Posted by **Anonymous** on Tuesday, May 18, 2010 at 9:17pm.

I actually have two questions:

4. An open box is to be made from a rectangular piece of material 3m by 2m by cutting a congruent square from each corner and folding up the sides. What are the dimensions of the box of the largest volume made this way, and what is the volume?

5. A cylindrical container w/ a circular base is to hold 64 cubic cm. Find its dimensions so that the amt (surface area) of metal required is a minimum when the container is

a. an open cup and

b. a closed can.

---

W/ 4, I have no idea how to approach. All I got is that volume of the box would be s(s-2)(s-3), I think.

As for 5, I know the formula for the volume and surface area, but my question is about the open cup for a. You probably have to subtract something from somewhere...but where? I don't really know.

Thank you very much!

- Calculus -
**Reiny**, Tuesday, May 18, 2010 at 9:30pm
4. Let each side of the square that is cut out be s m

then the box will have dimensions s by 2-2s by 3 - 2s

and the

volume = s(2-2s)(3-2s)

expand, find the derivative, set that equal to zero and solve that quadratic

This question is probably used more than any other to introduce the concept of maximum/minimum by most textbooks, only the numbers will differ.

5. another straight-forward Calculus question

let the radius be r, and the height h cm

volume = πr^2h

πr^2h = 64

h = 64/(πr^2)

a) SA = one circle + the "sleeve" of the cylinder

= πr^2 + 2πrh

= πr^2 + 2πr(64/(πr^2))

= πr^2 + 128/r

d(SA)/dr = 2πr - 128/r^2 = 0 for a max/min of SA

2πr = 128/r^2

r^3 = 64/π

take cube root, sub back into h = ....

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