Example  2
Alexander has a rectangular backyard hockey rink with an area of \(420 \thinspace \rm{ft}^2\). He wants to enclose all four sides with boards. Can he enclose the entire rink with \(80 \thinspace \rm{ft}\) of boards?
Step 1: Draw a diagram of the problem.

Let \(x\) represent the length of the rink.

Let \(y\) represent the width of the rink.

\(\begin{align}
 P &= 2l + 2w \\
 80 &= 2x + 2y \\
 80 - 2x &= 2y \\
 40 - x &= y \\
 \end{align}\)

Step 2: Build a quadratic equation to represent the problem.

\(\begin{align}
 {\rm{Area}} &= \thinspace {\rm{length }}\thinspace  \times \thinspace {\rm{ width}} \\
 420 &= x\left( {40 - x} \right) \\
 420 &= 40x - x^2 \\
 x^2 - 40x + 420 &= 0 \\
 \end{align}\)

Step 3: Analyze how to solve the problem.

If there is a Real solution to this problem, the discriminant will either be equal to \(0\) or greater than zero.

Step 4: Solve the problem.

\(a = 1, b = -40, c = 420\)

\(\begin{align}
 b^2 - 4ac &= \left( { - 40} \right)^2 - 4\left( 1 \right)\left( {420} \right) \\
  &= 1600 - 1680 \\
  &= -80  \end{align}\)

Step 5: Conclusion.

Because the discriminant is less than \(0\), there are no Real roots. This means that there is no solution to the problem and it is not possible for Alexander to enclose his \(420 \thinspace \rm{ft}^2\) backyard rink with \(80 \thinspace \rm{ft}\) of boards.




For further information about Using the Quadratic Formula see pp. 244 to 253 of Pre-Calculus 11.