# Math Help - optimization

1. ## optimization

You are supposed to design a rectangular box, open at the top and with a square base, that is to have a volume of exactly 647.5145 cm3. If the box is to require the least amount of material, what must be its dimensions?

2. "You are supposed to design a rectangular box, open at the top and with a square base, that is to have a volume of exactly 647.5145 cm3. If the box is to require the least amount of material, what must be its dimensions?"

First, we'll use the given info. to find some relation between the three dimensions. If z, y and x stand for the three dimensions length, breadth and height, we see that length and the breadth are the same as the base is a square.

So z=y.

The given info next is that the volume of the open box is 647.5145, we have xyz = 647.5145

Substituting z=y, we have xy^2 = 647.5145

The next is the function which represents the amount of material.
Hence, we gotta consider the surface area of a cuboid. We are also given that the top of the box is open, which means the area under consideration would be 2xz + 2xy + yz (we have only one yz because that represents the bottom face of the box, there is no top)

Using z=y, we have 2xy + 2xy + y^2 = 4xy + y^2

Now, we gotta use the relation xy^2 = 647.5145 in the expression 4xy + y^2 settle down for the function to be minimized either in 'x' alone or 'y' alone.

Once that is done, the rest is pretty simple,
i) equate the first derivative to zero to get a value for 'x' or 'y'.
ii) verify that the value of 'x' or 'y' found in step 1, when substituted in the second derivative gives a positive value.

This shows that the function is minimum for this value.

Find the other dimensions with the already given information.

Hope it helped,
MAX

3. ahh so helpful, thank you

4. yes, i understand now. thank you for walking me through it