Please show work. 1) A square bottomed box without a top has a fixed volume of 3

Question

Please show work.

1) A square bottomed box without a top has a fixed volume of 32 cm^3. What dimensions minimize the surface area.

2) To construct a closed tin can ( right circular cylinder) from 150picm^3 of material, what dimensions maximize volume?

3) An advertising sign is allowed to be 50sf, but it must be bordered on the top and bottom by 6" of blank area and on the left and right sides by 1' again for blank area to allow for installation, what dimensions allow the largest area inside these borders for the advertisement?

Explanation / Answer

(1) Let the box be A x A x B.

Then, A^2*B = 32

Surface area S = A^2 + 4AB = A^2 + 128/A

For minimum S, dS/dA = 0 => 2A - 128/A^2 = 0 => A^3 = 64 => A = 4 => B = 32/16 = 2

So, box should be 4 cm x 4 cm x 2 cm

(2)Let the height be H and radius of base be R.

Then, pi*R^2*H = 150*pi => R^2*H = 150

Surface Area S = 2*pi*R*H + 2*pi*R^2 = 300*pi/R + 2*pi*R^2

For maximum area, dS/dR = 0 => -150/R^2 +2R = 0 => R^3 = 75 => R = 4.217 cm

and, H = 8.434 cm

(3) Let the dimensions be A x B.

Then AB = 50

Area of advertisement is S = (A - 0.5 - 0.5)(B - 1 - 1) = (A-1)(B-2) = (A-1)(A/50 - 2) = A^2/50 - 2.02A + 2

For maximum area, dS/dA = 0 => A/25 = 2.02 => A = 50.5 ft

=> B = 0.99 ft

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