There are (at least) two different methods to solve this problem. One method uses calculus, so it would be the expected solution method for a calculus course. Another method doesn't use calculus but it does use knowledge of parabolas and quadratic expressions, so it would be the expected solution for a precalculus or algebra course. For the first part of the solution, both methods are the same.
Step 1 for this type of question is to always ALWAYS draw a diagram. Here's mine:
| | width w
| Maximize Area A |
Step 2 is to label the diagram with the information you know and what you're expected to find, and to assign variables. I did that in the diagram above.
Step 3 is to create equations for the information you have. Because the fence doesn't include the side by the river, we can't use the formula for the perimeter of a rectangle, 2l + 2w =116. Instead, the total amount of fencing is equal to two widths and one length:
l + 2w =116.
Our goal is to maximize the area. Here we can use the formula for the area of a rectangle:
Maximize A = lw.
Step 4 is to try to change the expression that you want to maximize so that it only uses one variable instead of two. We can most easily do this by solving the first equation for l and substituting the result into the equation for area.
Because l = 116 − 2w, we want to:
maximize A = lw = (116 − 2w)w . If we write this expression in the standard form for a quadratic, our goal is to:
maximize A = −2w2 + 116w.
Step 5 (for precalculus) is to recognize that the expression you want to maximize is a quadratic. This means that the function A(w) will take the shape of a parabola when A is represented on the y-axis and w is on the x-axis. Because the leading coefficient is negative, the parabola will point down with a maximum at its vertex.
You're expected to remember that the formula for the vertex of a parabola gives the location of the vertex as x = −b/(2a) when the quadratic is in standard form: y = ax2+bx+c. The expression we're interested in uses w instead of x as the variable, but the method is the same. With a=−2 and b=116,
w at the vertex where A is a maximum = −b/(2a) = −116/(2(−2)) = 29.
If you ever forget the formula for the vertex of a parabola, just remember it's the same as the quadratic formula when the discriminant b2-4ac is zero:
y=(−b±√discriminant)/(2a) = −b/(2a) when discriminant = b2-4ac =0.
The quadratic formula adds and subtracts the same value from the x-value of the vertex to determine where the curve crosses the x-axis.
Step 5 (for calculus) is to use the fact that the derivative of a function set equal to zero yields a critical point as its solution. We don't need to memorize any silly vertex formulas for calculus! For the function:
A = −2w2 + 116w, the derivative (using the power rule) set equal to zero is:
dA/dw = −4w + 116 = 0.
Solving this equation yields w = 29. This is the only critical point, so if we trust the person who wrote the question then it has to be a maximum. We could prove it's a maximum by using the 1st or 2nd derivative tests (which you can do on your own).
Step 6 (for either precalculus or calculus) is to determine the value of l by substituting back into the equaton for l from step 4. Try this step on your own.