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2 years ago

14

............please help

1 answer:

Novay_Z [31]2 years ago

6 0

X-intercept: (3.5,0)
Y-intercept: (2.5,0)

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A rectangular field with one side along a river is to be fenced. Suppose that no fence is needed along the river, the fence on t

Andreas93 [3]

Answer:

a) Side Parallel to the river: 200 ft

b) Each of the other sides: 400 ft

Step-by-step explanation:

Let L represent side parallel to the river and W represent width of fence.

The required fencing (F) would be $F=L+2W$ .

We have been given that field must contain 80,000 square feet. This means area of field must be equal to 80,000.

$LW=80,000...(1)$

We are told that the fence on the side opposite the river costs $20 per foot, and the fence on the other sides costs $5 per foot, so total cost (C) of fencing would be $C=20L+5(2W)\Rightarrow 20L+10W$ .

From equation (1), we will get:

$L=\frac{80,000}{W}$

Upon substituting this value in cost equation, we will get:

$C=20(\frac{80,000}{W})+10W$

$C=\frac{1600,000}{W}+10W$

$C=1600,000W^{-1}+10W$

To minimize the cost, we need to find critical points of the the derivative of cost function as:

$C'=-1600,000W^{-2}+10$

$-1600,000W^{-2}+10=0$

$-1600,000W^{-2}=-10$

$-\frac{1600,000}{W^2}=-10$

$-10W^2=-1,600,000$

$W^2=160,000$

$\sqrt{W^2}=\pm\sqrt{160,000}$

$W=\pm 400$

Since width cannot be negative, therefore, the width of the fencing would be 400 feet.

Now, we will find the 2nd derivative as:

$C''=-2(-1600,000)W^{-3}$

$C''=3200,000W^{-3}$

$C''=\frac{3200,000}{W^3}$

Now, we will substitute $W=400$ in 2nd derivative as:

$C''(400)=\frac{3200,000}{400^3}=\frac{3200,000}{64000000}=0.05$

Since 2nd derivative is positive at $W=400$ , therefore, width of 400 ft of the fencing will minimize the cost.

Upon substituting $W=400$ in $L=\frac{80,000}{W}$ , we will get:

$L=\frac{80,000}{400}\\\\L=200$

Therefore, the side parallel to the river will be 200 feet.

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3 years ago