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Your challenge is to create a cylindrical can that minimizes the cost of materials but must hold 100 cubic inches. The top and bottom of the can cost $0.014 per square inch, while the sides cost only $0.007 per square inch. Show how you did it too?

Sagot :

MrRoyal

Answer:

[tex]Radius = 1.997\ in[/tex] and [tex]Height = 7.987\ in[/tex]

[tex]Cost = \$1.05[/tex]

Step-by-step explanation:

Given

[tex]Volume = 100in^3[/tex]

[tex]Cost =\$0.014[/tex] -- Top and Bottom

[tex]Cost =\$0.007[/tex] --- Sides

Required

What dimension of the cylinder minimizes the cost

The volume (V) of a cylinder is:

[tex]V = \pi r^2h[/tex]

Substitute 100 for V

[tex]100 = \pi r^2h[/tex]

Make h the subject

[tex]h = \frac{100 }{\pi r^2}[/tex]

The surface area (A) of a cylinder is:

[tex]A = 2\pi r^2 + 2\pi rh[/tex]

Where

[tex]Top\ and\ bottom = 2\pi r^2[/tex]

[tex]Sides = 2\pi rh[/tex]

So, the cost of the surface area is:

[tex]C = 2\pi r^2 * 0.014+ 2\pi rh * 0.007[/tex]

[tex]C = 2\pi r(r * 0.014+ h * 0.007)[/tex]

[tex]C = 2\pi r(0.014r+ 0.007h)[/tex]

Substitute [tex]h = \frac{100 }{\pi r^2}[/tex]

[tex]C = 2\pi r(0.014r+ 0.007*\frac{100 }{\pi r^2})[/tex]

[tex]C = 2\pi r(0.014r+ \frac{0.007*100 }{\pi r^2})[/tex]

[tex]C = 2\pi r(0.014r+ \frac{0.7}{\pi r^2})[/tex]

[tex]C = 2\pi (0.014r^2+ \frac{0.7}{\pi r})[/tex]

Open bracket

[tex]C = 2\pi *0.014r^2+ 2\pi *\frac{0.7}{\pi r}[/tex]

[tex]C = 0.028\pi *r^2+ \frac{2\pi *0.7}{\pi r}[/tex]

[tex]C = 0.028\pi *r^2+ \frac{2 *0.7}{r}[/tex]

[tex]C = 0.028\pi *r^2+ \frac{1.4}{r}[/tex]

[tex]C = 0.028\pi r^2+ \frac{1.4}{r}[/tex]

To minimize, we differentiate C w.r.t r and set the result to 0

[tex]C' = 0.056\pi r - \frac{1.4}{r^2}[/tex]

Set to 0

[tex]0 = 0.056\pi r - \frac{1.4}{r^2}[/tex]

Collect Like Terms

[tex]0.056\pi r = \frac{1.4}{r^2}[/tex]

Cross Multiply

[tex]0.056\pi r *r^2= 1.4[/tex]

[tex]0.056\pi r^3= 1.4[/tex]

Make [tex]r^3[/tex] the subject

[tex]r^3= \frac{1.4}{0.056\pi }[/tex]

[tex]r^3= \frac{1.4}{0.056 * 3.14}[/tex]

[tex]r^3= \frac{1.4}{0.17584}[/tex]

[tex]r^3= 7.96178343949[/tex]

Take cube roots of both sides

[tex]r= \sqrt[3]{7.96178343949}[/tex]

[tex]r= 1.997[/tex]

Recall that:

[tex]h = \frac{100 }{\pi r^2}[/tex]

[tex]h = \frac{100 }{3.14 * 1.997^2}[/tex]

[tex]h = \frac{100 }{12.52}[/tex]

[tex]h = 7.987[/tex]

Hence, the dimensions that minimizes the cost are:

[tex]Radius = 1.997\ in[/tex] and [tex]Height = 7.987\ in[/tex]

To calculate the cost, we have:

[tex]C = 2\pi r(0.014r+ 0.007h)[/tex]

[tex]C = 2* 3.14 * 1.997 * (0.014*1.997+ 0.007*7.987)[/tex]

[tex]Cost = \$1.05[/tex]

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