Question

question 1 (5 points) a rectangular yard is enclosed by 100 meters of fencing. the table shows some possible values for the length and width of the yard. complete the table with the missing value of a, b, and c. a = 600 b = 15 c = 400 what are the dimensions that maximize the area of the yard? length = by width = blank 1: 600 blank 2: 15 blank 3: 400 blank 4: blank 5:

Explanation:

Step1: Recall perimeter formula

Let the length be $l$ and width be $w$. The perimeter of a rectangle with one - side not fenced (assuming one length side is against a wall or something) is $P = l + 2w$. Given $P=100$, so $l = 100 - 2w$. The area formula of a rectangle is $A=l\times w=(100 - 2w)w=100w-2w^{2}$.

Step2: For the table values

When $w = 15$, $l=100 - 2\times15=100 - 30 = 70$, and $A=l\times w=70\times15 = 1050$. But we are filling the table values.
When $l = 25$, from $l = 100 - 2w$, we solve for $w$: $25=100 - 2w$, then $2w=100 - 25 = 75$, and $w = 37.5$. And $A = 25\times37.5=937.5$.
When $w = 30$, $l=100 - 2\times30=100 - 60 = 40$, and $A=40\times30 = 1200$.
To find the maximum of the area function $A(w)=100w - 2w^{2}$, we can use the formula for the vertex of a parabola. For a quadratic function $y = ax^{2}+bx + c$, the $x$ - coordinate of the vertex is $x=-\frac{b}{2a}$. Here $a=-2$ and $b = 100$. So $w=-\frac{100}{2\times(-2)}=25$. Then $l=100-2\times25 = 50$, and $A = 50\times25=1250$.
For the values in the table:

• When length $l = 35$, $w=\frac{100 - 35}{2}=32.5$, and $A = 35\times32.5 = 1137.5$.
• When length $l = 40$, $w=\frac{100 - 40}{2}=30$, and $A = 40\times30=1200$.
• When width $w = 10$, $l=100 - 2\times10=80$, and $A = 80\times10 = 800$.

To maximize the area, we know from the vertex of the quadratic $A(w)=-2w^{2}+100w$, the width $w = 25$ meters, and the length $l=50$ meters.

Answer:

Blank 1: 50
Blank 2: 25
Blank 3: 1250
Blank 4: 37.5
Blank 5: 937.5