Question

A soft drink can is h centimeters tall and has a radius of r cm. The cost of material in the can is 0.0005 cents per cm2 and the soda itself costs 0.002 cents per cm3. The cans are currently 4 cm tall and have a radius of 4 cm. Use calculus to estimate the effect on costs of increasing the radius.

Answer 1

Answer:

Step-by-step explanation:

Given:

• The cost of material in the can is 0.0005 cents per cm2

• The soda itself costs 0.002 cents per cm3.

1. The Volume of the can is: V = Sh = π$r^{2}$ h = 3.14*$4^{2}$ *4 = 200.96 $cm^{3}$

2. The surface area of the can is

S = 2π$r^{2}$ + 2rhπ = 2*3.14*$4^{2}$ + 2*4*4*3.14

= 100.48 + 100.48

= 200.96 $cm^{2}$

The total cost of the can is:  0.0005*200.96$cm^{2}$  + 0.002*200.96 $cm^{3}$

If we increase the radius, the cost will be increased as well. Because the volume and the surface area will increase.

Answer 2

Answer:

Step-by-step explanation:

Given:

Cost of material, Cm = 0.0005 cents per cm2

Cost of soda, Cs = 0.002 cents per cm3

Height, h = 4 cm

Radius, r = 4 cm

Surface area of a cylinder, Ac = 2πrh + 2πr^2

= 2π × 4 × (4 + 4)

= 64π cm^2

Volume of a cylinder, V = πr^2 × h

= π × (4^2) × 4

= 64π cm^3

dAs = dAs/dr ×dr + dAs/dh × dh

= 2πh + 4πr × dr + 2πr × dh

= 8π + 16π × dr + 8π × dh

= 24π dr + 8π dh

dr = 0.1 cm

dh = -0.8 cm

dAs = 2.4π - 6.4π

= -4π × 0.0005

Cost = -0.00628

dV = dV/dr ×dr + dV/dh × dh

= 2πrh× dr + πr^2 × dh

= 32π × dr + 16π × dh

dr = 0.1 cm

dh = -0.8 cm

dAs = 3.2π - 12.8π

= -9.6π × 0.002

Cost = -0.603

Total cost = -0.00628 - 0.603

= -0.609