I am lost in this question, i made the triangle but cant keep going for the maximum. Need asap please.

Question

a rancher wants to use 100m of fence to construct a diagonal fence connecting two existing walls that meet a right angle. How should this be done so that the area enclosed by two walls and fence is a maximum?

Answer 1

This problem emphasized the concept of maximum.  The concept of minimum is also very important in calculus.  Be sure to get a good grasp of both concepts as soon as possible.
 
With two walls that are perpendicular (note:  this is a very good starting clue), you might (theoretically, for now) create a function A=f(P₁,P₂) that returns the area inside the right triangle P₁-P₂-C (where C is the corner point).  You already know that the area of a triangle is A=(1/2)bh, so the appropriate x and y values of the points are used to provide the result.
 
Thinking somewhat logically (since this is so-far theoretical), if the 100m of fence is very close to either side of the wall the enclosed area is only a little more than 0.  As P₁ moves along one side P₂ is forced to move along the other side (the maximum of 100m is certainly going to produce the maximum area, so why use less wire?).  Then, as the wire gets close to the other wall, the enclosed area gets closer to 0 again.  Someday, using calculus, you will get a formula that you can solve to show that the maximum area occurs when you have a 45º-45º-90º triangle.   [note: I use MS Excel, so I could put in lengths of 1, 2, 3, 4, 5, ... for one leg of the triangle and 100m for the hypotenuse and calculate the length of the other leg of the triangle using the Pythagorean Theorem.  Then, computing A=(1/2)bh is easy and I could plot the resulting graph.  You might have enough math to imagine that.]
 
Now, you have a diagram showing 100m as the hypotenuse of an isosceles right triangle.  From there, finding P₁ and P₂ along the walls for constructing the fence is easy.