a rectangle's area is length*width

total length =1400

2 equations
plug length equ into area equ

-----------

money= price* students
add a variable to account for 1 incremental change in price/students

(price - (2*variable) )
(students+variable*10)

This post was edited by saber_x3 on Jun 30 2014 07:22pm

For the second one:

Let:
r = revenue
p = price charged per student
n = number of students enrolled in the class

Hopefully it is intuitive that revenue is a function of price and enrollment:

We can also express the number of students enrolled as a function of price.


where (28-p) / 2 represents the number of decrease by $2

Substituting:



Expanding:



To find the maximum: Take the first derviative of the above expression and set it equal to 0.




If you don't undestand the derivative bit, you can always graph the function to see how price and revenue are interacting.

You know that Area = L * W

so A(x) = x(1400-2x)
= 1400x - 2x^2

You now have a quadratic, where a = -2 and b = 1400

Does this sound familiar?

Maximum area is where x = - b/2a
-1400 / 2(*-2)
350 ft


Plug back into our "Length" which we decided up there in the illustration is 1400 - 2x
1400 - 2(350)
700 ft


Max fence should be 700 ft (legnth) x 350 ft (width)

The amount of fenced in area would be the same no matter how you positioned the fencing. If you're looking for the most amount of space, that's a different question entirely.