Length= 5 dft¿¿¿
width =2 ft
Explanation
Step 1
Let
L represents the length
W represents the width
then
[tex]\text{Area}=L\cdot W[/tex]if the enclosure needs to have an area of at least 10 square feet, then
[tex]\begin{gathered} \text{Area}\ge10ft^2 \\ L\cdot W\ge10\text{ Inequality (1)} \end{gathered}[/tex]Step 2
The length of the enclosure needs to be 3 feet longer than the width, in other words you have to add 3ft to the width to obtain the length
[tex]L=3+W\text{ Equation (1)}[/tex]Step 3
replace equation(1) into inequality (1)
[tex]\begin{gathered} L\cdot W\ge10\text{ Inequality (1)} \\ (3+W)\cdot W\ge10 \\ 3W+W^2\ge10 \\ W^2+3W-10\ge0 \end{gathered}[/tex]now, we need to factorize, we need two numbers a and b, such
[tex]\begin{gathered} a+b=3 \\ a\cdot b=-10 \\ \text{then} \\ a=5 \\ b=-2 \end{gathered}[/tex]So,
[tex]\begin{gathered} W^2+3W-10\ge0 \\ (W+5)(W-2)\ge0 \\ \end{gathered}[/tex]the solutions are
[tex]\begin{gathered} W+5\ge0\text{ and W-2}\ge0 \\ W\ge-5\text{ and W}\ge2 \end{gathered}[/tex]we need a positive value for the width, so we take
[tex]W\ge2[/tex]Now, replace the value of W in equation (1) to find L
[tex]\begin{gathered} L=3+W\text{ Equation (1)} \\ L=3+2 \\ L=5\text{ ft} \end{gathered}[/tex]
