Question

The typical soft drink can in the U.S. has a volume of 355 cm3. The two circular ends cost $0.0006 per cm2 each (because they are thicker), and the cost of the aluminum for the side of the can is $0.00026 per cm2. What dimensions will minimize the cost of a can? (Round your answers to two decimal places.)

Answer 1

solution is attached below

Answer 2

Answer:

Radius = 4.25 cm

Height = 6.26 cm

C = $0 05 per can

Step-by-step explanation:

Volume of a typical soft drink in U.S = 355cm^3

The cost of the two circular ends per cm^2 = $0.0006

The cost of aluminum for the side of the can per cm^2 = $0.00026

Total surface area of can(A)

= 2πr^2 + 2πrh

Volume (V) = πr^2h

πr^2h = 355

h = 355/ πr^2

The cost function = C

C = 0.0006(2πr^2) + 0.00026(2πrh)

C = 0.0012πr^2 + 0.00052(πr*355/πr^2)

C = 0.0012πr^2 + 0.1846/r

To minimize cost, differentiate C with respect to r

dC/dr = 2(0.0012πr) - 0.1846/r^2

= 0.0024πr - 0.1846/r^2

To minimize cost, dC/dr = 0

0.0024πr - 0.1846/r^2 = 0

0.0024πr = 0.1846/r^2

(0.0024πr) r^2 = 0.1846

0.0024πr^3 = 0.1846

r^3 = 0.1846/0.0024

r^3 = 76.91667

r = cuberoot if 76.91667

r = 4.25 cm

Recall that h = 355/πr^2

h = 355 / π(4.25)^2

h = 6.26 cm

Recall that

C = 0.0012πr^2 + 0.1846/r

C = 0.00012π(4.25)^2 + 0.1846/4.25

C = 0.0068094 + 0.043435

C = 0.050244

C = $ 0.05 per can (approximate to 2 d.p)