Question

among all rectangles that have a perimeter of 8 ft, find the dimensions of the one with the largest area. length width area ft ft ft2 resources read it -/3 points 0/30 submissions used details my notes

Explanation:

Step1: Set up perimeter formula

Let the length of the rectangle be $l$ and the width be $w$. The perimeter formula is $P = 2(l + w)$. Given $P=8$, so $2(l + w)=8$, which simplifies to $l + w=4$, and we can express $l = 4 - w$.

Step2: Set up area formula

The area formula of a rectangle is $A=l\times w$. Substitute $l = 4 - w$ into the area formula, we get $A=(4 - w)w=4w - w^{2}$.

Step3: Find the maximum of the area function

Since $A(w)=4w - w^{2}$ is a quadratic function of the form $y = ax^{2}+bx + c$ with $a=- 1$, $b = 4$, $c = 0$. The vertex of a quadratic function $y = ax^{2}+bx + c$ has its $x$ - coordinate (in our case $w$ - coordinate) at $w=-\frac{b}{2a}$. Substituting $a=-1$ and $b = 4$ into $w=-\frac{b}{2a}$, we have $w=-\frac{4}{2\times(-1)} = 2$.

Step4: Find the length

Since $l + w=4$ and $w = 2$, then $l=4 - 2=2$.

Answer:

length: 2 ft, width: 2 ft, area: 4 ft²