A boy wants to build a rectangular pen that is made out of steel on two(opposite) sides and aluminum on the other two sides for his science project. The steel fencing costs $18 per linear foot and...

A boy wants to build a rectangular pen that is made out of steel on two(opposite) sides and aluminum on the other two sides for his science project. The steel fencing costs $18 per linear foot and the aluminum fencing is $13 per linear foot. The pen needs to contain 200 square feet of space.

a. Let x be the length of one of the aluminum sides. Find a function for the cost of the pen in terms of x

b. Find the dimensions of the pen that will minimize the cost (to the nearest tenth if needed).

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lemjay | High School Teacher | (Level 3) Senior Educator

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(a)
First, apply the formula of area of rectangle to express the dimension of the side made out of steel in terms of x.

`A= l e n g t h * width`

Then, plug-in length=x and A=200.

`200= x* width`

And, divide both sides by x to isolate width.

`200/x = width`

Hence, the sides of the rectangle are x (aluminum) and 200/x (steel).

So, to get the cost function of the pen, multiply each sides by of the rectangle by the unit price of the material.

`C= 13x+13x+18*200/x+18*200/x`

`C=26x+7200/x`

Hence, the cost function of the pen in terms of x is `C(x) =26x+7200/x ` .

(b) To solve, take the derivative of the cost function.

`C'(x) = (26x+7200/x)' = 26 - 7200x^(-2)`

`C'(x)=26-7200/x^2`

Then, set C'(x) equal to zero.

`0=26-7200/x^2`

To simplify, multiply both sides by x^2.

`0=26x^2-7200`

Then, use the quadratic formula to solve for x.

`x=(-b+-sqrt(b^2-4ac))/(2a)=(-0+-sqrt(0^2-4(26)(-7200)))/(2*26)`

`x=(+-sqrt748800)/52`

`x=+-16.64`

Since x represents the length of the side made of aluminum, take only the positive value. So,

`x=16.6`

Now that the value of x is known, determine the length of the side made of steel. To do so, plug-in the value of x to:

`200/x=200/16.6 =12.0`

Hence, the dimension of the rectangular pen that will minimize the cost is `16.6 xx 12.0` feet.

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