mcr3u unit 1 test due: september 24th @10:30am app /18 t&i /23 7. ms. smith has 190 m of fencing to enclose…

Explanation:

Step1: Let width be $w$ and length be $l$.

The perimeter of rectangle is $2(l + w)$. Given perimeter $2(l + w)=190$, so $l + w = 95$, then $l=95 - w$.

Step2: Write area function.

The area $A$ of a rectangle is $A=l\times w$. Substitute $l = 95 - w$ into the area formula, we get $A(w)=(95 - w)w=95w - w^{2}$.

Step3: Find domain.

Since width $w>0$ and length $l = 95 - w>0$ (dimensions must be non - negative), solving $95 - w>0$ gives $w < 95$. So the domain is $0<w<95$.

Step4: Find range.

The function $A(w)=-w^{2}+95w$ is a quadratic function with $a=-1$, $b = 95$, $c = 0$. The vertex of a quadratic function $y = ax^{2}+bx + c$ has $x$ - coordinate $x=-\frac{b}{2a}$. Here $w=-\frac{95}{2\times(-1)} = 47.5$. Substitute $w = 47.5$ into $A(w)$: $A(47.5)=-(47.5)^{2}+95\times47.5=47.5\times(95 - 47.5)=47.5\times47.5 = 2256.25$. Since $a=-1<0$, the parabola opens downwards. So the range is $0<A(w)\leq2256.25$.

Step5: Find maximum - area dimensions.

As found in Step4, when $w = 47.5$, $l=95 - 47.5=47.5$.

Answer:

a) $A(w)=95w - w^{2}$ b) Domain: $(0,95)$; Range: $(0,2256.25]$ c) Width = 47.5 m, Length = 47.5 m