Respuesta :
Given the perimeter and area of a rectangle you have to determine its possible width and length.
The perimeter of the rectangle can be calculated as:
[tex]\begin{gathered} P=2w+2l \\ 72=2w+2l \end{gathered}[/tex]The area of the rectangle can be calculated as:
[tex]\begin{gathered} A=wl \\ 288=wl \end{gathered}[/tex]With this we have determined an equation system:
[tex]\begin{gathered} 72=2w+2l \\ 288=wl \end{gathered}[/tex]First step: write the first equation in terms of the length:
[tex]\begin{gathered} 72-2w=2l \\ l=\frac{72}{2}-\frac{2w}{2} \\ l=36-w \end{gathered}[/tex]Second step: replace the expression obtained in the second formula:
[tex]\begin{gathered} 288=wl \\ 288=w(36-w) \end{gathered}[/tex]Third step solve the term in parentheses by applying the distributive property of multiplication
[tex]\begin{gathered} 288=36\cdot w-w\cdot w \\ 288=36w-w^2 \end{gathered}[/tex]Fourth step, equal to zero and solve using the quadratic formula:
[tex]-w^2+36w-288=0[/tex]This is a quadratic expression where
a=-1
b=36
c=-288
The quadratic formula is
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]Replace with the coefficients to calculate the possible values of the width
[tex]\begin{gathered} w=\frac{-36\pm\sqrt[]{(36)^2-4(-1)(-288)}}{2(-1)} \\ w=\frac{-36\pm\sqrt[]{1296-1152}}{-2} \\ w=\frac{-36\pm\sqrt[]{144}}{-2} \\ w=\frac{-36\pm12}{-2} \end{gathered}[/tex]Fifth step, calculate both possible values for w:
Positive:
[tex]\begin{gathered} w_1=\frac{-36+12}{-2} \\ w_1=12ft \end{gathered}[/tex]Negative:
[tex]\begin{gathered} w_2=\frac{-36-12}{-2} \\ w_2=24ft \end{gathered}[/tex]So the possivle values for the width are:
w₁=12ft
w₂=24ft
With this, calculate the possible lengths
Length one:
[tex]\begin{gathered} l_1=36-w_1 \\ l_1=36-12 \\ l_1=24ft \end{gathered}[/tex]Length two:
[tex]\begin{gathered} l_2=36-w_2 \\ l_2=36-24 \\ l_2=12ft \end{gathered}[/tex]So the possible values of width and length of the rectangle are:
w₁=12ft, l₁=24ft
w₂=24ft, l₂=12ft
