Question

suppose you are designing a rectangular garden and you have 24 meters of fencing with which to enclose the garden. a. can you enclose a garden that is 7 meters long and 5 meters wide with the available fencing? what is the area of the garden? b. can you enclose a garden that is 9 meters long and 3 meters wide with the available fencing? what is the area of the garden? c. by computing the area of other possible gardens that can be enclosed with 24 meters of fencing, find or estimate the dimensions of the garden that has the most area. a. a garden that is 7 meters long and 5 meters wide can/cannot be enclosed with the available fencing. the area of the garden is square meters.

Explanation:

Step1: Calculate perimeter of 7 - m long and 5 - m wide garden

Perimeter of a rectangle $P = 2(l + w)$, where $l = 7$m and $w = 5$m. So $P=2(7 + 5)=2\times12 = 24$m.

Step2: Calculate area of 7 - m long and 5 - m wide garden

Area of a rectangle $A=l\times w$, so $A = 7\times5=35$m².

Step3: Calculate perimeter of 9 - m long and 3 - m wide garden

$P = 2(9 + 3)=2\times12 = 24$m.

Step4: Calculate area of 9 - m long and 3 - m wide garden

$A=9\times3 = 27$m².

Step5: Find maximum - area rectangle

For a rectangle with perimeter $P = 24$m, $P = 2(l + w)=24$, so $l + w=12$, and $w = 12 - l$. Then $A=l\times w=l(12 - l)=12l - l^{2}$. This is a quadratic function. The vertex of the quadratic function $y = ax^{2}+bx + c$ (here $a=-1$, $b = 12$, $c = 0$) has $x=-\frac{b}{2a}$. So $l=-\frac{12}{2\times(-1)} = 6$m. When $l = 6$m, $w = 6$m (a square), and $A = 6\times6=36$m².

Answer:

a. A garden that is 7 meters long and 5 meters wide can be enclosed with the available fencing. The area of the garden is 35 square meters.
b. A garden that is 9 meters long and 3 meters wide can be enclosed with the available fencing. The area of the garden is 27 square meters.
c. The dimensions of the garden with the most area are 6 meters by 6 meters, and the area is 36 square meters.