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
[tex]h = \frac{120}{r^2}[/tex]
The base and sides will cost $.01 per cm² to produce
Total cost of the base and side [tex]c_1[/tex] = 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 [tex]c_2[/tex] = 0.02 ( πr²)
Overall Total cost = [tex]c_1 + c_2[/tex]
= 0.01 ( πr² + 2πrh) + 0.02 ( πr²)
= 0.01 πr² + 0.02 πrh + 0.02 πr²
= [tex]0.03 \pi r^2 + 0.02 \pi r (\frac{120}{r^2} )[/tex]
= [tex]0.03 \pi r^2 + 2.4 \frac{\pi}{r}[/tex]
Taking the differentiation to find the radius dimension to minimize cost; we have:
[tex]\frac{dc}{dr} =0[/tex] ⇒ [tex]0.06 \pi r^2 - \frac{2.4 \pi}{r^2} =0[/tex]
[tex]0.06 \pi r^2 = \frac{2.4 \pi}{r^2}[/tex]
[tex]r^4 = \frac{2.4 \pi}{0.06 \pi}[/tex]
[tex]r^4 = 40[/tex]
[tex]r = \sqrt[4]{40}[/tex]
[tex]r= 2.515[/tex] cm
However, [tex]\frac{d^2c}{dr^2}=0.12 \pi r + \frac{4.8 \pi}{r^3}[/tex]
[tex]\frac{d^2c}{dr^2}|__{r= 2.515}} =0.12 \pi (2.515) + \frac{4.8 \pi}{(2.515)^3} >0[/tex]
Therefore; we can say that the cost is minimum at r = 2.515 since it is positive.
To determine the height ; we have:
[tex]h = \frac{120}{r^2} \\h = \frac{120}{(2.515)^2}[/tex]
[tex]h = \frac{120}{6.34}[/tex]
h = 18.93 cm
