You have 600 feet of fencing to enclose a rectangular plot that borders on a river If you do not fence the side along the river find the length and width of the plot that will maximize the area?

300

150

This is going to be a long one...

So you want to know the area, so we can start off with the formula for area:

A = L * W

From here, we can start plugging in numbers. We can say that the two sides that are perpendicular to the river are the width (W), and the side parallel to the river is the length (L). We know that there are two W sides, and one L side, and all three of them add up to 600.

We are going to utilize a System of Equations

Here is our second equation:

600 = 2W + L

Now that we've got that, we need some way to combine them. So let's solve for L (in the second equation since it has numbers) and plug it into the area formula.

600 = 2W + L

subtract 600 from both sides

0 = 2W + L -600

subtract L from both sides

-L = 2W -600

divide each side by -1

L = -2W + 600

I divided by -1 to get just L.

Now, we can plug that into the formula for area, like I said above.

A = W * (-2W + 600)

From here, we can distribute the W to the -2w+600.

A = -2w2 + 600w

Now we can factor out -2

A = -2(w2 - 300w)

Obviously this is a quadratic equation, and we can solve it by Completing the Square.

To make this easier to understand, I'll add a C term (0).

A = -2(w2 - 300w) + 0

In order to complete the square, we need to take the B term (-300) and take half of it, and then square that. Then we will have to add that inside the parentheses, but we will need to subtract the same number to balance the equation. Because, of course, we can't just subtract a random number from the equation! In order to balance the equation, we will take the C term (which I added above) and subtract the "magic" square multiplied by the A term (-2).

(... and THAT'S how you complete the square) :D

Reading that over, it sounds ridiculously complicated, but hey, it's not like I'm getting paid to write this. Hopefully writing it out will help. Otherwise, just believe me.

A = -2(w2 -300w) + 0

Now we are going to complete the square.

We shall take -300 and divide it by 2 (take half of it)

Then square -150 (^which is obviously half of -300^)

and end up with 22500 (oh no big number!)

This number will be added inside of the ()

A = -2(w2 - 300w + 22500) +0

But, obviously, we cant just add 22500 out of nowhere, we need to "counter" it to keep things balanced.

So we will (do something pretty complicated) take the "A" term (-2, no not the A at the other side of the equation) and multiply it by 22500 and subtract it from C.

So 22500 * -2 is -45000 and -45000 minus 0 is 45000

A = -2(w2 - 300w +22500) + 45000

Now that we've got that "completing the square" junk out of the way, we can easily factor it.

A = -2(w-150)2 +45000

This is the standard form of a quadratic equation. (150, 45000) is the vertex. This means that when the width is 150, the area will be 45000.

So what does that mean?

Well if we look, our "A" term once again, not the A at the beginning of the equation, which stands for area! is -2. That means that are graph will be a "unhappy parabola" aka it will look like a ∩. This means that our vertex will be the highest point on the graph, meaning it has the highest area.

Continuing on, we know that the ideal width is 150, but we still need to find the idea length. To do so, we will plug 150 into our original equation that we solved for L. (such a long time ago)

Here's the equation we made of L earlier:

L = -2W + 600

Now, since we know w (which, yes, is the same as capital W, I just made it lowercase to make it easier to see when it was squared) we can plug it in!

L = -2(150) + 600

Now we can simplify.

L = -300 + 600

ADD!

L = 300

So now we have our final answer (finally)

LENGTH: 300

WIDTH: 150

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