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omeli [17]
3 years ago
5

HELP ASAP PLLLLLLLLLLS!

Mathematics
1 answer:
PolarNik [594]3 years ago
8 0

Answer:

uummmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm

Step-by-step explanation:

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Please hawlp pls.......
Amiraneli [1.4K]

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.

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Please graph as this coordinate plane is. Thank you
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Answer:

hope this helps :)

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Step-by-step explanation:

Because they are all different numbers

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