SOLUTION: 1. A rectangular field is to be enclosed by 600m of fence. What dimensions will give the maximum area? What is the maximum area?
2. The captain of a riverboat charges $36 pe
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Question 204444: 1. A rectangular field is to be enclosed by 600m of fence. What dimensions will give the maximum area? What is the maximum area?
2. The captain of a riverboat charges $36 per person including lunch. The cruise averages 300 customers a day. The captain is considering increasing the price. A survey of the customers indicates that for every $2 increase, there would be 10 fewer customers. What increase in price would maximize the revenue?
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
The perimeter of a rectangle is given by:
which can be restated as:
The area of a rectangle is given by:
Substituting:
 = \left(\frac{P - 2w}{2}\right)w = \frac{Pw - 2w^2}{2})
This function has an extremum where the first derivative is equal to zero:
Setting the first derivative equal to zero:
This extremum is a maximum if the value of the 2nd derivative is negative.
Which is negative for all values of the independent variable, hence the extremum is a maximum.
In order for the width to be one fourth of the perimeter, the length must also be one-fourth of the perimeter. That is to say, the maximum area for a rectangle with a given perimeter is achieved when the rectangle is a square with side measure of one-fourth of the perimeter.
=====================================================
Let 
represent the price of one fare. Then the revenue function is the number of passengers times the price of a fare. The price of one fare minus 36, quantity divided by 2 is the number of $2 increases to the price. So:
 = \left(300 - 10\left(\frac{p-36}{2}\right)\right)p)
Which simplifies to:
First derivative:
Set equal to zero:
Second derivative:
Hence extreme point is a maximum.
John
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