Probem13+Jimmy,+Sylvia

Heeeeeey JIMMY! Please check this stuff. I might be super wrong. Don't worry. I checked it and we got the same answer. Unless we both made the same mistakes, we're fine.

Sylvia: First, Given: A box = 1200 cm^2

So. Here is our box: We know the total area of this box is first the area of the bottom, then 4 times the area of one side (to account for all 4 sides) A box = (b * b) + 4(h * b) A box = b^2 + 4hb = 1200 Solve for h 4hb = 1200 - b^2 hb = 300 - b^2/4 h = 300/b - b/4

We know Volume = L * W * H V = b * b * h V = b^2 h

Substitute h for 300/b - b/4 V = b^2 (300/b - b/4)

Simplify 300b - (b^3)/4

Derive! dV/db = 300 - 3/4 (b^2) dV/db = -.75b^2 + 300

Critical points -.75b^2 + 300 = 0 .75b^2 = 300 b^2 = 400 b = +/- 20 (but you know it's not -20 because we don't want negative volume)

V = 300b - (b^3)/4 V = 300(20) - (20^3)/4 V = 6000 - 2000 V = 4000 cm^3

I do agree with most of this, except that all you showed is that we have a local max at the critical value -- how can we be sure that the behavior at the endpoints isn't going to upset the proverbial apple cart and give us some bigger values? Remember the routine for finding the //ABSOLUTE// extrema!