Question

Please hawlp pls.......

Answer 1

(a)
`A(x)=160x-2x^(2)`

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

(c) Maximum area
`=3200` square meters

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,
`(dA)/(dx)=0\\(d)/(dx)(160x-2x^(2))=0\\160-4x=0\\4x=160\\x=(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* 1600\\A(x=40)=6400-3200=3200\textrm{ }m^(2)`

Therefore, maximum area is 3200 square meters.