optimization 10.3

Question

Drum Tight Containers is designing an​ open-top, square-based, rectangular box that will have a volume of 1687.5 in.^3. What dimensions will minimize surface​ area? What is the minimum surface​ area?

What are the dimensions of the​ box?

The length of one side of the base is ____

The height of the box is ___

Answer 1

cube root of 1687.5= 11.9055 inches = edge length of the cube which minimizes surface area

if all 6 sides are closed

height = about 11.9 inches

width = length = about 11.9 inches

minimum surface area = 6(11.9055)^2 = about 850.45 square inches

but that's if it's not open topped.

an open top makes the surface area smaller

Volume = V = 1687.5= Lwd = dw^2

d= 1687.5/w^2

Surface Area = A = 4dw +w^2= 4(1687.5)/w + w^2

A' = 2w - 6750/w^2 = 0

w-3375/w^2 = 0

w^3 = 3375

w = 15 inches wide

L=15 inches Long

d = 1687.5/225 = 7.5 inches deep or high

minimum Surface area = 15^2 +4(7.5)(15) = 225+450= 675 square inches

Answer 2

Let x be the length of a side of the square base. Let h be the height.

Volume, V= x²h = 1687.5

h = 1687.5/x²

Surface Area, S = x² + 4xh = x²+4(1687.5)/x

S'(x) = 2x - 4(1687.5)/x² = (2x³-4(1687.5))/x²

To find the minimum, set S'(x) to zero and solve for x

x = (2(1697.5))^1/3 = 15 in

h = 1697.5/15² = 7.5 in

Surface Area, S= 15²+4(15)(7.5) = 675 in²

Dimensions of the box = 15×15×7.5

Checking if minimum with second derivative test:

S''(15) = 2-4(1687.5)(-2)/15³ >0, so x = 15 is the minimum.

Answer 3

Surface area s = 4xy + x²; volume v = x²y = 1687.5; y = 1687.5/x²; so, s(x) = 6750/x + x².

s'(x) = - 6750/x² + 2x = 0; x³ = 3375; x = 15 in; y = 1687.5/225 = 7.5 in

s''(x) = 13500/x³ + 2; s''(15) = 13500/3375 + 2 =6 > 0. So, at x = 15, y = 7.5 we have minimum

s_min = s(15) = 6750/15 + 225 = 450 + 225 = 675 in²