Answer:
(a) [tex]A(x)=160x-2x^{2}[/tex]
(b) [tex]x=40\textrm{ m}[/tex]
(c) Maximum area[tex]=3200[/tex] square meters
Step-by-step explanation:
(a)
Given:
Total length for fencing = 160 m
Width of the rectangle = [tex]x[/tex] m.
Let the other length of the rectangle be [tex]y[/tex] m.
Then, from the figure,
[tex]x+y+x=160\\2x+y=160\\y=160-2x[/tex]
Now, area of a rectangle is equal to the product of its length and width.
So, area is, [tex]A(x)=xy=x(160-2x)=160x-2x^{2}[/tex]
(b)
Given:
[tex]A(x)=160x-2x^{2}[/tex]
For maximum area, derivative of area with respect to [tex]x[/tex] must be 0.
So, [tex]\frac{dA}{dx}=0\\\frac{d}{dx}(160x-2x^{2})=0\\160-4x=0\\4x=160\\x=\frac{160}{4}=40\textrm{ m}[/tex]
Therefore, for maximum area, [tex]x=40\textrm{ m}[/tex].
(c)
Given:
[tex]A(x)=160x-2x^{2}[/tex]
Maximum area occurs at [tex]x=40[/tex]. Plug in 40 for [tex]x[/tex] in [tex]A(x)[/tex] expression. This gives,
Maximum area is,
[tex]A(x=40)=160(40)-2(40)^{2}\\A(x=40)=6400-2\times 1600\\A(x=40)=6400-3200=3200\textrm{ }m^{2}[/tex]
Therefore, maximum area is 3200 square meters.
