Respuesta :
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 [tex]F=L+2W[/tex].
We have been given that field must contain 80,000 square feet. This means area of field must be equal to 80,000.
[tex]LW=80,000...(1)[/tex]
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 [tex]C=20L+5(2W)\Rightarrow 20L+10W[/tex].
From equation (1), we will get:
[tex]L=\frac{80,000}{W}[/tex]
Upon substituting this value in cost equation, we will get:
[tex]C=20(\frac{80,000}{W})+10W[/tex]
[tex]C=\frac{1600,000}{W}+10W[/tex]
[tex]C=1600,000W^{-1}+10W[/tex]
To minimize the cost, we need to find critical points of the the derivative of cost function as:
[tex]C'=-1600,000W^{-2}+10[/tex]
[tex]-1600,000W^{-2}+10=0[/tex]
[tex]-1600,000W^{-2}=-10[/tex]
[tex]-\frac{1600,000}{W^2}=-10[/tex]
[tex]-10W^2=-1,600,000[/tex]
[tex]W^2=160,000[/tex]
[tex]\sqrt{W^2}=\pm\sqrt{160,000}[/tex]
[tex]W=\pm 400[/tex]
Since width cannot be negative, therefore, the width of the fencing would be 400 feet.
Now, we will find the 2nd derivative as:
[tex]C''=-2(-1600,000)W^{-3}[/tex]
[tex]C''=3200,000W^{-3}[/tex]
[tex]C''=\frac{3200,000}{W^3}[/tex]
Now, we will substitute [tex]W=400[/tex] in 2nd derivative as:
[tex]C''(400)=\frac{3200,000}{400^3}=\frac{3200,000}{64000000}=0.05[/tex]
Since 2nd derivative is positive at [tex]W=400[/tex], therefore, width of 400 ft of the fencing will minimize the cost.
Upon substituting [tex]W=400[/tex] in [tex]L=\frac{80,000}{W}[/tex], we will get:
[tex]L=\frac{80,000}{400}\\\\L=200[/tex]
Therefore, the side parallel to the river will be 200 feet.
