of all rectangles with a perimeter of 32, which one has the maximum area? if x and y are the length and…

of all rectangles with a perimeter of 32, which one has the maximum area? if x and y are the length and width of the rectangle, respectively, then the area of the rectangle is a = xy, where 2x + 2y = . writing the area function as a function of x, it follows that the area is a(x)= , where ≤x≤ . evaluate a at the endpoints of and at the critical point of a. it follows that a has an absolute maximum value at x = . therefore, the rectangle that has the maximum area has a length of and a width of . (simplify your answers.)

of all rectangles with a perimeter of 32, which one has the maximum area? if x and y are the length and width of the rectangle, respectively, then the area of the rectangle is a = xy, where 2x + 2y = . writing the area function as a function of x, it follows that the area is a(x)= , where ≤x≤ . evaluate a at the endpoints of and at the critical point of a. it follows that a has an absolute maximum value at x = . therefore, the rectangle that has the maximum area has a length of and a width of . (simplify your answers.)

Answer

Explanation:

Step1: Find perimeter - related equation

Given perimeter $P = 32$, and $P=2x + 2y$, so $2x + 2y=32$, which simplifies to $y = 16 - x$.

Step2: Express area as a function of x

The area $A=xy$. Substitute $y = 16 - x$ into it, we get $A(x)=x(16 - x)=16x - x^{2}$.

Step3: Determine the domain of x

Since $x>0$ and $y = 16 - x>0$, then $0<x<16$.

Step4: Find the critical - point

Differentiate $A(x)$ with respect to $x$. $A^\prime(x)=\frac{d}{dx}(16x - x^{2})=16 - 2x$. Set $A^\prime(x)=0$, then $16 - 2x = 0$, solving for $x$ gives $x = 8$.

Step5: Evaluate the area function

Evaluate $A(x)$ at the endpoints $x = 0$ and $x = 16$ and the critical - point $x = 8$. $A(0)=0$, $A(16)=0$, $A(8)=8\times(16 - 8)=64$.

Answer:

If $x$ and $y$ are the length and width of the rectangle, respectively, then the area of the rectangle is $A = xy$, where $2x + 2y = 32$. Writing the area function as a function of $x$, it follows that the area is $A(x)=16x - x^{2}$, where $0<x<16$. Evaluate $A$ at the endpoints of $[0,16]$ and at the critical point of $A$. It follows that $A$ has an absolute maximum value at $x = 8$. Therefore, the rectangle that has the maximum area has a length of $8$ and a width of $8$.