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A pencil cup with a capacity of 45 in.3 is to be constructed in the shape of a rectangular box with a square base and an open top. If the material for the sides costs 27¢/in.2 and the material for the base costs 90¢/in.2, what should the dimensions of the cup be to minimize the construction cost?

Answer :

david8644

Answer:

A 3x3 inches base.

A 3x5 inches sides

Explanation:

As they want the rectangular prism with a square base, you need to make a 3 by 3 base so it is a square and then you just add a 5x3 side to make it 45 in3 and that creates you the minimal cost, because the cheapest would be to make a 1 by 1 but in that one you couldnt fit the pencils efficciently.

Lanuel

To minimize the construction cost, the dimensions of the cup be 3 x 3 x 5 inches.

Given the following data:

  • Capacity = 45 [tex]in^3[/tex].
  • Cost of material (sides) =  27 ¢[tex]/in^2[/tex]
  • Cost of material (base) =  90 ¢[tex]/in^2[/tex]

How to calculate the dimensions of the cup.

Let the height be h.

Let the square base be x.

Therefore, the capacity (volume) of the pencil cup is given by:

[tex]V=x^2h\\\\45=x^2h\\\\h=\frac{45}{x^2}[/tex]

At the minimum cost, we have:

[tex]C(x)=90x^2+27(4xh)\\\\C(x)=90x^2+108xh\\\\C(x)=90x^2+108x(\frac{45}{x^2} )\\\\C(x)=90x^2+\frac{4860}{x}[/tex]

Differentiating wrt x, we have:

[tex]C(x)=180x-\frac{4860}{x^2} =0\\\\180x^3-4860=0\\\\180x^3=4860\\\\x^3=27\\\\x=\sqrt[3]{27}[/tex]

x = 3 inches.

For the height:

[tex]h=\frac{45}{x^2} \\\\h=\frac{45}{3^2}\\\\h=\frac{45}{9}[/tex]

h = 5 inches.

Dimensions = [tex]square \;base \times height[/tex]

Dimensions = 3 x 3 x 5 inches.

Read more on capacity here: brainly.com/question/25248189

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