Answer
Step-by-step explanation
Variables
• x: width of the rectangle, in ft
,• y: length of the rectangle, in ft
Given that the length, y, is 30 feet less than 6 times the width, x, then:
[tex]y=6x-30\text{ \lparen eq. 1\rparen}[/tex]The area of a rectangle is calculated as follows:
[tex]Area=length{}t\times width[/tex]In this case, the area is 4836 square ft. Substituting this value and using the before defined variables, we get:
[tex]4836=yx\text{ \lparen eq. 2\rparen}[/tex]Substituting equation 1 into equation 2:
[tex]\begin{gathered} 4836=(6x-30)x \\ 4,836=6x^2-30x \\ 0=6x^2-30x-4836 \end{gathered}[/tex]We can solve this equation with the help of the quadratic formula, as follows:
[tex]\begin{gathered} x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ x_{1,2}=\frac{3^0\pm\sqrt{(-30)^2-4\cdot6\cdot(-4836)}}{2\cdot6}\frac{}{} \\ x_{1,2}=\frac{30\pm\sqrt{116964}}{12} \\ x_{1,2}=\frac{30\pm342}{12} \\ x_1=\frac{30+342}{12}=31 \\ x_2=\frac{30-342}{12}=-26 \end{gathered}[/tex]Given that the width cannot be negative, then the second solution is discarded.
Substituting x = 31 ft into equation 1:
