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?


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,


{{{(2x+2y)/2=40/2=20}}}


{{{x+y=20}}}


{{{y=20-x}}}


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


3.   Substitute y in Step 1 into Step 2.


{{{xy=x(20-x)=51}}}  Multiplying the terms will yield:  {{{20x-x^2=51}}}


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.


 {{{20x-x^2-20x+x^2= 51+x^2-20x}}}


Simplifying will yield


{{{0 = 51+x^2-20x}}}


Finally, the above equation is equivalent to


{{{x^2-20x+51=0}}}


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


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 


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


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


*[invoke quadratic "x", 1, -20, 51 ]



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.


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