Hi Adam,
Let's see if we can figure this "bad boy" out.
Let's denote the edges on the base of the box, which is square, by b, and the height of the box by h
Surface Area. The box has no top. So the surface area is the area of the base (b2) plus the area of all four of the sides. Since the area of one side is bh, it follows that the area of all four sides is 4hb. Therefore, the total surface area is
S = b2+4hb
and since the surface are is 240 cm2, we have that
240cm2 = b2+4hb
Solving for h, we have
h = (240-b2)/(4b)
So, this gives us a relation for h in terms of b. We can now turn to the volume.
V = b2h.
Substituting for h, we have
V = b2(240-b2)/4b = 60b - b3/4
Taking the derivative of V with respect to be we have
dV/db = 60 - 3b2/4
Setting that to 0 and solving for b, we find that b is equal to the square root of 80
b = √(80)
You can evaluate the second derivative to see if this is a max or a min. Doing so reveals that
d2V/db2 = -1.5√80
So, since the second derivative is negative, we have a maximum.
We know what b is. So, all we need to do is find out the value for h. To do this, we return to our surface area equation, which we have solved for h
h = (240-b2)/(4b)
Now, we substitute √80 for b
h = (240-b2)/(4b) = (240-(√80)2)/(4√80) = (240-80)/(4√80) = 160/(4√80) =40/√80
Note that we do not like to have radicals in the denominator, so we will multiply both the numerator an denominator by √80
h= 40√80/80 = √80/2 or h=(1/2)√80
So, the dimensions of the box are b=√80 cm and h = (1/2)√80 cm, and the volume is 40√80 cm3
Hope you found this helpful. Please feel free to contact me if you need help with anything else.
Thanks,
Bob