John has 120 feet of fencing to make a kennel for his dog

Question

He decides to make it in the shape of a rectangle, using his house as one side of the kennel. What are the dimensions John should use so that his dog has the maximum space to move around?

Answer 1

For this problem, you're trying to maximize the area of a rectangle which is given by A = length * width.

Since his house is one length of the rectangle, let's call that the width, w.

I will note that it seems like they should have told you that his house is at least some feet wide so that you don't end up making a recommendation for using the house as the width and finding out that his house is only 20 feet wide!

The length of fence would thus be length + length + width = 120

And your area is length * width = Area

I always recommend you try guessing and checking to see how this might start fitting together. If you do that, you see that it's not as simple as, "longer length equals more area" or "wider width equals more area," the quantities are not related linearly, meaning we are dealing with an optimization problem that involves quadratics.

Perhaps I could have made an equation to help me out, then graphed it and found the max. If you look, there is a system of equations that we've started to develop: 2l + w = 120 and w*l = Area, so let's solve w in terms of length and optimize that.

You should get w = 120 - 2l, which can then be plugged in, yielding 120(length) -2(length)² which we can now plug in to a graph and find the max. If you don't have a graphing calculator handy you could use a derivative to optimize: I've subbed in x for l, giving me this function to optimize: f(x) = 120x-2x²

Optimizing tells us to take the derivative and set it equal to zero to find the min/max, and in our case it yields our length, which we can then use to find our width from our original equation. That should take you the rest of the way to the answer!

Answer 2

Let's take sides of the rectangle as a and b.

The area of the rectangle is then a*b

We know there is only 120 ft of fence, so

120 ft = a + 2*b

Let's solve for a and plug it into the area equation:

a = 120 - 2b

Area = a*b = (120 - 2b)*b

Area = 120b - 2b²

Now we want to maximize this equation, so take the derivative and set it equal to zero:

Area' = 0 = 120 - 4b

solving for b,

b = 30 ft

Now we can plug this back into the fence equation to get a:

120 ft = a + 2*(30 ft)

a = 60 ft

So he should have one side be 60 ft and two sides be 30 ft, and then his house would need to be 60 ft to accommodate the 4th side of the rectangle.

Let's check:

Our area is 60 ft * 30 ft = 1800 ft²

Take one case where you take a little bit from the 60 ft side and lengthen the 30 ft sides:

New dimensions are 59 ft and 30.5 ft

Area = 59 ft * 30.5 ft = 1799.5 ft² (less, so that's good)

Take another case where you take a little bit from the 30 ft sides and lengthen the 60 ft side:

New dimensions are 61 ft and 29.5 ft

Area = 61 ft * 29.5 ft = 1799.5 ft² (less, so that's good)

We verified our maximum.