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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 .
We have been given that field must contain 80,000 square feet. This means area of field must be equal to 80,000.
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 .
From equation (1), we will get:
Upon substituting this value in cost equation, we will get:
To minimize the cost, we need to find critical points of the the derivative of cost function as:
Since width cannot be negative, therefore, the width of the fencing would be 400 feet.
Now, we will find the 2nd derivative as:
Now, we will substitute in 2nd derivative as:
Since 2nd derivative is positive at , therefore, width of 400 ft of the fencing will minimize the cost.
Upon substituting in
, we will get:
Therefore, the side parallel to the river will be 200 feet.
Answer: A.
The inverse sine function is written as sin−1(x) or arcsin(x). Inverse functions swap x- and y-values, so the range of inverse sine is − π/2 to π/2 and the domain is −1 to 1. When evaluating problems, use identities or start from the inside function.
(REFER TO CHART BELOW)
The inverse sine function is written as sin−1(x) or arcsin(x). Inverse functions swap x- and y-values, so the range of inverse sine is − π/2 to π/2 and the domain is −1 to 1. When evaluating problems, use identities or start from the inside function.
(REFER TO CHART BELOW)