Help Please!!!!!!!!!!
A three sided fence is to be built next to a straight river which forms the fourth side of a rectangular region. The enclosed area is to equal 3200 ft2. The cost of the fence is $2/ft. This question will have you find the dimensions that minimize the cost of the fence.
(a) What quantity needs to be optimized?
(b) Draw a picture modeling this situation.
40
(c) Find a function for the quantity that is being optimized and find its domain.
(d) Find the dimensions that minimize the cost of the fence.

2. The perimeter of the fence is x+y+x = 2x+y ft (fourth side is river). The total cost of the fence is $2(2x+y). This quantity needs to be optimized.

3. Note that x>0 and y>0. See attached diagram for graphs.

5. Since x·y=3200, then

[tex]y=\dfrac{3200}{x}[/tex]

and function

[tex]f(x)=2(2x+y)=4x+2y=4x+\dfrac{6400}{x}[/tex]

is the function for the quantity that is being optimized.

The domain of this function is [tex]x\neq 0.[/tex]

6. Find the derivative f'(x):

[tex]f'(x)=4-\dfrac{6400}{x^2}.[/tex]

When f'(x)=0, you have

[tex]4-\dfrac{6400}{x^2}=0,\\ \\4x^2-6400=0,\\ \\x^2-1600=0,\\ \\x^2=1600,\\ \\x_1=40,\ x_2=-40.[/tex]

The width of the fence must have positive length, then x=40 ft and y=80 ft.

MrRoyal MrRoyal · Answer 2 · 2021-11-10T09:02:53+01:00

The area of the three sided fence is the product of its dimensions.

(a) The quantity that needs to be optimized is the cost of the perimeter of the fence

(b) See attachment for the picture of the model

(c) The function

The area is given as:

[tex]\mathbf{A = 3200}[/tex]

From the figure, the perimeter is given as:

[tex]\mathbf{P = 2x + y}[/tex]

So, the area is:

[tex]\mathbf{A = xy = 3200}[/tex]

Make y the subject

[tex]\mathbf{y = \frac{3200}x}[/tex]

Substitute [tex]\mathbf{y = \frac{3200}x}[/tex] in [tex]\mathbf{P = 2x + y}[/tex]

[tex]\mathbf{P = 2x + \frac{3200}{x}}[/tex]

The fence costs $2 per 1 ft.

So, we have:

[tex]\mathbf{P = 2 \times (2x + \frac{3200}{x})}[/tex]

[tex]\mathbf{P = 4x + \frac{6400}{x}}[/tex]

So, the function to optimize is:

[tex]\mathbf{P(x) = 4x + \frac{6400}{x}}[/tex]

And the domain is:

[tex]\mathbf{x >0}[/tex]

(d) The dimension that optimizes the fence

We have:

[tex]\mathbf{P(x) = 4x + \frac{6400}{x}}[/tex]

Differentiate

[tex]\mathbf{P'(x) = 4 - \frac{6400}{x^2}}\\[/tex]

Set to 0

[tex]\mathbf{4 - \frac{6400}{x^2} = 0}[/tex]

Rewrite as:

[tex]\mathbf{4 = \frac{6400}{x^2} }[/tex]

Cross multiply

[tex]\mathbf{4x^2 = 6400}[/tex]

Divide both sides by 4

[tex]\mathbf{x^2 = 1600}[/tex]

Take positive square roots

[tex]\mathbf{x = 40}[/tex]

Substitute [tex]\mathbf{x = 40}[/tex] in [tex]\mathbf{y = \frac{3200}x}[/tex]

[tex]\mathbf{y = \frac{3200}{40}}[/tex]

[tex]\mathbf{y = 80}[/tex]

Hence, the dimension that minimizes the cost of the fence is 40 ft by 80 ft

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