If you have 266 meters of fencing to enclose a rectangular field what is the maximum area that can be enclosed?

Here you are given a perimeter and told that the field is rectangular. That gives us one equation. 2L + 2W = 266 (A rectangle has two pairs of sides - L and W - that are equal. The sum of all sides is equal to the perimeter by definition.) Area is calculated by multiplying length and width. This gives us our second formula. L * W = Area In the past, in algebra we could solve for one of the variables and then substitute and solve for the other. This is trickier. We CAN solve for one of the variable, for example, for L 2L + 2W = 266 2L = 266 - 2W L = 133 - W When we substitute that in to the other equation, however, we seem to hit a roadblock. (133 - W) * W = Area 133W - W2 = Area Now we still have two variables. but if you think back to the problem, there are multiple possible answers depending on the length and widths that are chosen. In this problem we want to maximize area. So instead, let's think of this problem as finding a maximum point of the following function: f(x) = 133W - W2 To find the rate at which the function (area), we take the derivitive of the function. f'(x) = 133 - 2W When a function reaches its maximum (or minimum point) the rate at which the value of the function is increasing (or decreasing) will go to zero. To better understand this, think of a rollercoaster. As your car climbs up a hill, the extra height you gain is still increasing, but near the top it will slowly level off to where you are not getting any higher (zero) and then decrease after. The same is true for this function. So what does that all mean? By setting the derivitive to zero, we can solve for the point(s) at which this function stops increasing (or decreasing) and hits a maximum (or minimum). f'(x) = 133 - 2W 0 = 133 - 2W 133 = 2W W = 133/2 or 66.5 The big question now is whether this is a minimum point in the function or a maximum. By taking the second derivitive we can test where this part of the function is concave up (like a U) or concave down (like an upside-down U). f''(x) = -2 This means that our function is concave down at all values of X. As the image suggests we are looking at a relative maximum point which is what we are looking for! So going back to our finding, we know that at the maximum area, W = 66.5; now we can solve for Y. 2L + 2W = 266 2L + 2(66.5) = 266 2L + 133 = 266 2L = 133 L = 66.5 So now we know the dimension of our rectangular field is 66.5 x 66.5. Now calculate the area with L x W. L x W = Area 66.5 x 66.5 = 4422.25 The maximum rectangular area that can be enclosed with 266 meters of Fencing is 4,422.25 square meters. Double check my work. I have to quick finish this so I cannot check my calculations!