To minimize cost, is to reduce cost to the barest minimum.
The dimensions that minimize the cost of a can are: radius of 2.90 cm and a height of 39.02 cm
Given
[tex]V = 355cm^3[/tex] --- volume of the can
[tex]C_1 = \$0.0006 cm^{-2}[/tex] --- the cost of the circular ends
[tex]C_2 = \$0.00026 cm^{-2}[/tex] --- the cost of the side
The soft drink can, has the shape of a cylinder.
The volume of the can is:
[tex]V = \pi r^2h[/tex]
So, we have:
[tex]\pi r^2h = 355[/tex]
Make h the subject
[tex]h =\frac{355}{\pi r^2}[/tex]
The surface area of the can is:
[tex]A = 2\pi rh + 2\pi r^2[/tex]
The cost of the can is then calculated as:
[tex]A =C_2 \times 2\pi rh + 2\pi r^2 \times C_1[/tex]
Substitute [tex]h =\frac{355}{\pi r^2}[/tex]
[tex]A =C_2 \times 2\pi r \times \frac{355}{\pi r^2} + 2\pi r^2 \times C_1[/tex]
[tex]A =C_2 \times 2 \times \frac{355}{r} + 2\pi r^2 \times C_1[/tex]
Substitute [tex]C_1 = \$0.0006 cm^{-2}[/tex] and [tex]C_2 = \$0.00026 cm^{-2}[/tex]
[tex]A =0.00026 \times 2 \times \frac{355}{r} + 2\pi r^2 \times 0.0006[/tex]
[tex]A =\frac{0.1846}{r} + 0.0012\pi r^2[/tex]
[tex]A =\frac{0.1846}{r} + 0.0038 r^2[/tex]
Rewrite as:
[tex]A =0.1846r^{-1} + 0.0038 r^2[/tex]
Differentiate
[tex]A =-1 \times 0.1846r^{-2} + 2 \times 0.0038 r[/tex]
[tex]A =-0.1846r^{-2} + 0.0076 r[/tex]
Set to 0
[tex]-0.1846r^{-2} + 0.0076 r = 0[/tex]
Rewrite as:
[tex]0.0076r = 0.1846r^{-2}[/tex]
Multiply through by [tex]r^2[/tex]
[tex]0.0076r^3 = 0.1846[/tex]
Solve for [tex]r^3[/tex]
[tex]r^3 = \frac{0.1846}{0.0076}[/tex]
[tex]r^3 = 24.2894[/tex]
Take cube roots of both sides
[tex]r = \sqrt[3]{24.2894}[/tex]
[tex]r = 2.90[/tex]
Recall that:
[tex]h =\frac{355}{\pi r^2}[/tex]
[tex]h = \frac{355}{\pi \times 2.90}[/tex]
[tex]h = 39.02[/tex]
Hence, the dimensions that minimize the cost of a can are:
[tex]r = 2.90[/tex] and [tex]h = 39.02[/tex]
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