2. the area of a rectangle is at most 360 square inches. The length is 6 more than twice the width. Find the length and width of the Rectangle.
A.How would you represent the dimensions of the rectangle?
B.What mathematical sentence would represent the given Situation?
C.what are the possible dimensions of the Rectangle? show your solution

Problem: The area of a rectangle is at most 360 square inches. The length is 6 more than twice the width. Find the length and width of the Rectangle.

A. How would you represent the dimensions of the rectangle?

Let l and w as the length and width of the rectangle respectively. And the formula of the area of a rectangle is Area = l • w. And substitute all the values.

B. What mathematical sentence would represent the given Situation?

I set into two equations to solve for length and the width.

[tex] \begin{cases} l = 6 + 2w \\ 360 =l⋅w \end{cases}[/tex]

C. What are the possible dimensions of the Rectangle? show your solution.

Solution:

[tex] \begin{cases} l = 6 + 2w& \green{(eq. \: 1) }\\ 360 =l⋅w& \green{(eq. \: 2)}\end{cases}[/tex]

- Substitute l from the first equation in terms of w.

[tex] \begin{cases} l = 6 + 2w \\ 360 =(6 + 2w)⋅w \end{cases}[/tex]

[tex] \begin{cases} l = 6 + 2w \\ 360 =2 {w}^{2} + 6w \end{cases}[/tex]

[tex] \begin{cases} l = 6 + 2w \\ 2 {w}^{2} + 6w - 360 = 0 \end{cases}[/tex]

- Solve the quadratic equation from the second equation by using the quadratic formula. Make sure that gives the positive solution.

[tex] \begin{gathered}w = \frac{ \text -6 + \sqrt{ {( \text6)}^{2} - 4(2)(\text-360)} }{2( \text - 2)} \end{gathered}[/tex]

[tex] \begin{gathered}w = \frac{ \text - 6 + \sqrt{36 - 4( \text -720)} }{4} \end{gathered}[/tex]

[tex] \begin{gathered}w = \frac{ \text - 6 + \sqrt{36 - 4( \text -720)} }{4} \end{gathered}[/tex]

[tex]\begin{gathered}w = \frac{ \text - 6 + \sqrt{36 - ( \text -2880)} }{4} \end{gathered}[/tex]

[tex]\begin{gathered}w = \frac{ \text - 6 + \sqrt{36 + 2880} }{4} \end{gathered}[/tex]

[tex]\begin{gathered}w = \frac{ \text - 6 + \sqrt{2916} }{4} \end{gathered}[/tex]

[tex]\begin{gathered}w = \frac{ \text - 6 + 54 }{4} \end{gathered}[/tex]

[tex]\begin{gathered}w = \frac{48}{4} \end{gathered}[/tex]

[tex]w = 12[/tex]

- Substitute w to the first equation to find the length.

[tex]\begin{cases} l = 6 + 2(12) \\ w = 12\end{cases}[/tex]

[tex]\begin{cases} l = 6 + 24 \\ w = 12\end{cases}[/tex]

[tex]\begin{cases} l = 30 \\ w = 12\end{cases}[/tex]

- The dimensions of the rectangle are:

[tex]\boxed{ \sf{ \: length = \green{30 \:inches }\: }}[/tex]

[tex]\boxed{ \sf{ \: width = \green{12 \:inches }\: }}[/tex]

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