#### If 1200 cm^{2} of material is available to make a box with a square base and an open top, find the largest possible volume of the box.

#### Solution:

Let “b” represent the length of the base and “h” represent the height. The surface area of the box can be represented as

1200=b^{2}+4hband isolating for “h” gives us,

h=\frac{(1200-b^{2})}{(4b)}

The volume is V=b^{2}h=\frac{b^{2}(1200-b^{2})}{(4b)}=300b-\frac{b^{3}}{4}

Thus,

V'(b)=300-\frac{3}{4}b^{2}

Now, we see where the absolute maximum is for our function by equating V'(b)=0.

0=300-\frac{3}{4}b^{2}300=\frac{3}{4}b^{2}

b^{2}=400

b=20.

Therefore, the absolute maximum is at b=20. Substitute this value back into h=\frac{(1200-b^{2})}{(4b)}

h=\frac{(1200-20^{2})}{(420)}h=10

Thus, the largest possible volume is b^{2}h=(20)^{2}(10)=4000cm^{3}

The height of the open box should be what?

10 cm, it’s actually at the bottom of the page 🙂