find the dimensions that minimize the surface area

Question

Waste Management Company wants to construct a dumpster in the shape of a rectangular solid with no top whose length is four times its width. Its volume is fixed at 25.6 yd³. Find the dimensions of the dumpster that will minimize its surface area.

Answer 1

Let x = width, y = length and z= height

Volume = xyz = 25.6 yd³

y = 4x

z = 25.6/(xy) = 25.6/(4x²) = 6.4/x²

Surface Area of sides and bottom, S = 2xz + 2yz + xy

S(x) = 2x(6.4/x²) + 2(4x)(6.4/x²)+4x² = 12.8/x + 51.2/x + 4x² = 64/x+4x²

Find S'(x), set to zero and solve for x.

S'(x) = -64/x²+8x

Multiply through by x²

-64+8x³ = 0

x = (8)^1/3 = 2 yds

y = 4x = 8 yds

z = 6.4/4 = 1.6 yds

S(1) = 68

S(2) = 48

S(3) = 57.33

So, x=2 is a minimum

Answer 2

v= Lwh = Length times width times height, L=4w

v= (4w)wh= 4hw^2

25.6 = 4hw^2

h = 25.6/4w^2= 6.4/w^2

surface A = 4 sides plus bottom area

=2hw + 2hL + Lw

= 2hw + 2h(4w) + 4w^2

= h(2w +8w) +4w^2

= (6.4/w^2)(2w+8w) + 4w^2

= (6.4/w^2)(10w) + 4w^2

= 64/w + 4w^2

take the derivative with respect to w

and set it equal to zero, solve for w

A'(w) = 8w - 64/w^2 = 0

= 8w^3 -64 = 0

8w^3=64

w^3 = 8

w = 2 yards wide

L =4w = 8 yards long

h =6.4/w^2 = 6.4/4 = 1.6 yards high

v = wLh = 2(8)1.6 = 16(1.6 = 25.6

surface A = h(10w) + 4w^2

=1.6(20) + 4(4)

= 32+16

= 48 square yards = minimum surface Area