Question

The base and sides will cost $.01 per cm2 to produce but the top, which is plastic and resealable, will cost $.02 per cm2 to produce. What should the dimensions be to minimize cost?

Answer 1

Here is the full question

A snack food company wishes to have a cylinder package for it's almond and cashew mix. The cylinder must contain 120 cm³ worth of product. The base and sides will cost $.01 per cm2 to produce but the top, which is plastic and resealable, will cost $.02 per cm2 to produce. What should the dimensions be to minimize cost?

Answer:

The radius and height are both dimension in the cylinder; in order to minimize the cost

radius = 2.515 cm

height = 18.93 cm

Step-by-step explanation:

We denote the radius of the cylinder to be = r

and the height of the cylinder = h

The volume of a cylinder is known to be = πr²h

Also, from the question; we are also told that the cylinder contains 120 πcm³

i.e πr²h = 120π

Dividing both sides with π; we have:

r²h = 120

$h = \frac{120}{r^2}$

The base and sides will cost $.01 per cm² to produce

Total cost of the base and side $c_1$ = 0.01 ( πr² + 2πrh)

but the top, which is plastic and resealable, will cost $.02 per cm² to produce.

i.e

cost of the top cylinder $c_2$ = 0.02 ( πr²)

Overall Total cost = $c_1 + c_2$

= 0.01 ( πr² + 2πrh) + 0.02 ( πr²)

= 0.01 πr² + 0.02 πrh + 0.02 πr²

= $0.03 \pi r^2 + 0.02 \pi r (\frac{120}{r^2} )$

= $0.03 \pi r^2 + 2.4 \frac{\pi}{r}$

Taking the differentiation to find the radius dimension to minimize cost; we have:

$\frac{dc}{dr} =0$ ⇒ $0.06 \pi r^2 - \frac{2.4 \pi}{r^2} =0$

$0.06 \pi r^2 = \frac{2.4 \pi}{r^2}$

$r^4 = \frac{2.4 \pi}{0.06 \pi}$

$r^4 = 40$

$r = \sqrt[4]{40}$

$r= 2.515$ cm

However, $\frac{d^2c}{dr^2}=0.12 \pi r + \frac{4.8 \pi}{r^3}$

$\frac{d^2c}{dr^2}|__{r= 2.515}} =0.12 \pi (2.515) + \frac{4.8 \pi}{(2.515)^3} >0$

Therefore; we can say that the cost is minimum at r = 2.515 since it is positive.

To determine the height ; we have:

$h = \frac{120}{r^2} \\h = \frac{120}{(2.515)^2}$

$h = \frac{120}{6.34}$

h = 18.93 cm