Palo alto college is planning to construct a rectangular parking lot on land bordered on one side by a highway. the plan is to use 760 feet of fencing to fence off the other three sides. what dimensions should the lot have if the enclosed area is to be a​ maximum

Width = 190 
Length = 380  
Let's first create an equation that expresses the area of the parking lot based upon it's width. 
The area is expressed as:
 A = W*L 
And the length is expressed as: 
L = 760 - 2*W 
So substituting the expression for L into the equation for A, gives: 
A = W*L 
A = W*(760 - 2*W) 
A = 760W - 2*W^2  
Since the area equation is a quadratic equation, the maximum value will happen on its axis of symmetry. So let's use the quadratic formula to get the roots in order to find the axis of symmetry. The roots are x = 0, and x = 380. So the axis of symmetry is x = (0 + 380)/2 = 190. So the ideal width is 190 feet and the length will be 760 - 190*2 = 760 - 380 = 380.  
Let's prove that 190 is the ideal width. I'll use the equation for area and the value (190 + e) as the width, then evaluate the expression. 
A = 760W - 2*W^2
 A = 760(190 + e) - 2*(190 + e)^2 
A = (144400 + 760e) - 2*(36100 + 380e + e^2)
 A = 144400 + 760e - (72200 + 760e + 2e^2)
 A = 144400 + 760e - 72200 - 760e - 2e^2
 A = 72200 - 2e^2
 
 Notice that the only term containing e is -2e^2. That value will be 0 if e = 0, and if e is any non-zero value, it will be negative and decrease the available area of 72200 square feet. Therefore the ideal width is 190 feet.