Answer:
the dimensions for the can that will minimize production cost is 9.13 cents
Step-by-step explanation:
The volume of a cylinder V = π r²h
If we make the height h the subject of the formula; we have :
h = V/ π r²
Given that the volume of the cylinder = 400
Then
h = 400/ π r²
The total cost will be: 0.02 × 2πrh + 0.07 × 2πr²
= 0.04 (πrh) + 0.14 (πr²)
= 0.04 (πr[tex]\frac{400} {\pi r^2}[/tex]) + 0.14 (πr²)
= 16/r + 0.14 (πr²)
total cost(c)= 16/r + 0.14 (πr²)
(c') = -16/r² + 0.28 (πr)
Let differentiate (c') with respect to zero (0); then:
-16/r² = - 0.28 (πr)
r³ = 16/0.28 π
r³ = 18.19
r = 2.63 cm
Recall that:
h = 400/ π r²
h = 400/ π (2.63)²
h = 400/21.73
h = 18.41 cm
From; total cost = 0.04 (πrh) + 0.14 (πr²)
replacing the value of r and h ; we have:
= 0.04 (π×2.63×18.41) + 0.14 (π × 2.63²)
= 0.04 (152.11) + 0.14 ( 21.73)
= 6.0844 + 3.0422
= 9.1266
≅ 9.13 cents
Therefore; the dimensions for the can that will minimize production cost is 9.13 cents
