Question

Please hawlp pls.......

Answer 1

Answer:

(a) $A(x)=160x-2x^{2}$

(b) $x=40\textrm{ m}$

(c) Maximum area$=3200$ square meters

Step-by-step explanation:

(a)

Given:

Total length for fencing = 160 m

Width of the rectangle = $x$ m.

Let the other length of the rectangle be $y$ m.

Then, from the figure,

$x+y+x=160\\2x+y=160\\y=160-2x$

Now, area of a rectangle is equal to the product of its length and width.

So, area is, $A(x)=xy=x(160-2x)=160x-2x^{2}$

(b)

Given:

$A(x)=160x-2x^{2}$

For maximum area, derivative of area with respect to $x$ must be 0.

So, $\frac{dA}{dx}=0\\\frac{d}{dx}(160x-2x^{2})=0\\160-4x=0\\4x=160\\x=\frac{160}{4}=40\textrm{ m}$

Therefore, for maximum area, $x=40\textrm{ m}$.

(c)

Given:

$A(x)=160x-2x^{2}$

Maximum area occurs at $x=40$. Plug in 40 for $x$ in $A(x)$ expression. This gives,

Maximum area is,

$A(x=40)=160(40)-2(40)^{2}\\A(x=40)=6400-2\times 1600\\A(x=40)=6400-3200=3200\textrm{ }m^{2}$

Therefore, maximum area is 3200 square meters.