SOLUTION: a farmer wants to set up a pigpen using 40 feet of fence to enclose a rectangular area of 51 square feet. what are the dimensions of the pigpen?

Question 211995: a farmer wants to set up a pigpen using 40 feet of fence to enclose a rectangular area of 51 square feet. what are the dimensions of the pigpen?
Answer by drj(1380) (Show Source): You can put this solution on YOUR website!
A farmer wants to set up a pigpen using 40 feet of fence to enclose a rectangular area of 51 square feet. what are the dimensions of the pigpen?

1. Perimeter of a rectangle is 2x+2y=40 where x=one side of rectangle and y=the adjacent side of x. We can simplify this as x+y=20 where we divided 2 on both sides of the equation. That is,

2. Area=xy=51 where area of rectangle is height times width.

3. Substitute y in Step 1 into Step 2.

Multiplying the terms will yield:

4. Put everything on the left side to the right. That is, add -20x+x^2 from both sides of the equation to make the left side equal to zero.

Simplifying will yield

Finally, the above equation is equivalent to

4. Now this is just a Quadratic Equation so we can use the formula

where a=1, b=-20 and c=51

5. Use the following steps to solve the quadratic equation.

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=196 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 17, 3. Here's your graph:

6. The rectangular sides are 3 and 17. As a check 3*17=51 (Area is 51 square feet) and Perimeter is 2*(3+17)=40 ft. So everything checks out.

For Step-by-Step videos in Introduction to Algebra, please visit http://www.FreedomUniversity.Tv/courses/IntroAlgebra
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