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mote1985 [20]
3 years ago
7

A beach has to enclose a rectangular area, because some endangered species are nesting there. They have 200 feet of rope to rope

off the area with. What is the maximum area that they can rope off?
Mathematics
1 answer:
kkurt [141]3 years ago
6 0
Area is equal to length times width. The perimeter (the amount of rope) has to equal twice the length added to twice the width so we're left with:
A = l * w
200 = 2l + 2w
solve for either l or w
l = 100 - w
plug into the area equation to get one equation with two variables
A = w(100 - w)
A = -w^2 + 100w
take the derivative
A' = -2w + 100
set the derivative equal to zero
0 = -2w + 100
2w = 100
w = 50
This is the width that maximizes the area
with a width of 50, the length must also be 50 to have a perimeter of 200
therefore, they can rope up to 50 * 50 = 2500 ft^2
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horrorfan [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

 g(h(x)) = \sqrt[4]{x^2 + x +5} + 3

b

 h(g(x))  = \sqrt{x}  + 7\sqrt[4]{x} + 17

c

 h(h(x)) =  [x^2 + x + 5 ]^2 + x^2 + x + 10

Step-by-step explanation:

From the question we are told that

    h(x) =  x^2  + x  + 5

and  

    g(x) = \sqrt[4]{x} + 3

Considering first question

Now we are told  g(h(x))

i.e

        g(h(x)) =  [x^2 + x + 5 ]^{\frac{1}{4} } + 3

=>     g(h(x)) = \sqrt[4]{x^2 + x +5} + 3

Considering second  question

   Now we are told  h(g(x))

i.e

    h(g(x)) =  [x^{\frac{1}{4} } + 3]^2 +  x^{\frac{1}{4} } + 3 + 5

=> h(g(x)) =  x^{\frac{1}{2} } + 6x^{\frac{1}{4} } + 9+ x^{\frac{1}{4}}  + 8

=>  h(g(x))  = x^{\frac{1}{2}} + 7x^{\frac{1}{4}} + 17

=>  h(g(x))  = \sqrt{x}  + 7\sqrt[4]{x} + 17

Considering third question

         h(h(x))= [x^2 + x + 5]^2 + [x^2 + x + 5 ] +  5

=>       h(h(x)) =  [x^2 + x + 5 ]^2 + x^2 + x + 10

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marishachu [46]
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