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

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> 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      Log On


   



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) About Me  (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:



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.

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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:



Which simplifies to:



First derivative:



Set equal to zero:



Second derivative:



Hence extreme point is a maximum.

John