Maximum dimensions

Question

The owner of the Rancho Los Feliz has 3120 yd of fencing to enclose a rectangular piece of grazing land along the straight portion of a river and then subdivide it by means of a fence running parallel to the sides. No fencing is required along the river. (See the figure below.) What are the dimensions of the largest area that can be enclosed?
x = ___ yd
y = ___ yd

What is this area?___ yd²

 

Equation for fence is 3x+y=3120

Answer 1

The nice thing about this problem is that they already gave you the fence equation, 3x + y = 3120, so the hard part is done. Let me explain quickly why it is 3x and not 4x though, because that is where most people lose this problem. There is no fencing along the river, so you only need one of the two sides that run parallel to the river, which is the y piece. Then you need two x pieces for the ends, plus one more x piece in the middle for the fence that subdivides it. Two ends plus the divider is three x pieces, so the total fencing is 3x + y, and that has to equal 3120.

Now we want to maximize the area, and area of a rectangle is just length times width, so A = xy. The problem is that A has two variables in it, and we can only work with one, so we use the fence equation to get rid of y. Solving 3x + y = 3120 for y gives us y = 3120 - 3x, and substituting that in:

A = x(3120 - 3x) = 3120x - 3x²

Remember, our purpose is to find the x that makes this as big as possible. Since this is a calculus class, we take the derivative and set it to zero:

A' = 3120 - 6x

3120 - 6x = 0, so 6x = 3120, which gives x = 520

Then we plug that back into the fence equation to get y:

y = 3120 - 3(520) = 3120 - 1560 = 1560

And the area is:

A = (520)(1560) = 811,200

To be sure this is a maximum and not a minimum, take the second derivative, which is A'' = -6. Since that is negative, the curve opens downward, so x = 520 really is the maximum. You can also just notice that A = 3120x - 3x² is a downward parabola, so its vertex has to be the highest point.

That brings up something worth knowing, which is that you did not actually need calculus for this one. Since the area came out to be a quadratic, you could find the vertex with x = -b/2a from Algebra 2, which here is -3120/(2)(-3) = 520. Same answer. I like checking optimization problems that way when the function turns out to be a quadratic, because if the two methods disagree then you know you made an algebra mistake somewhere while setting it up.

I also checked a few values on either side, and the areas come out to 810,900 at x = 510 and 810,900 again at x = 530, both of which are smaller than 811,200, so 520 really is the peak. That symmetry around the answer is another sign the setup was right.

So, our final answer is x = 520 yd, y = 1560 yd, and the largest area is 811,200 yd².