Question

A rancher plans to make four identical and adjacent rectangular pens against a​ barn, each with an area of 400 msquared. What are the dimensions of each pen that minimize the amount of fence that must be​ used?

Answer 1

Answer:

base = $10\sqrt{5}m=22.36m$ and height = $8\sqrt{5}m = 17.89m$

Step-by-step explanation:

In order to solve this problem, we must first understand how the pens are going to be placed. Refer to the attached diagram for an image of this.

From this diagram, we can start building the equations we are going to use to solve this. Since the problem doesn't specify the length of the barn, we will have no restriction on this. Taking this into account, we can start by building an equation for the area of each pen.

The area of a rectangle is given by the equation:

$A=bh$

where b is the base of the rectangle and h is the height of the rectangle.

with the given information we can now build our first equation:

$400=bh$

Since we have an equation with two variables, or two unknowns, we must have a second equation we can use to find the b and h values we need. In order to get this second equation, we must take into account that we need to minimize the amount of fencing to be used, so we can build an equation that will represent this fencing. When looking at the diagram, we can see that the fencing will be used to cover 4 bases and 5 heights, so our fencing equation should look like this:

$F=4b+5h$

Where F represents the amount of fencing to be used.

So now we can solve the first equation to substitute it into the second. Let's solve for the base b.

$bh=400\\\\b=\frac{400}{h}$

Which can now be substituted into the second equation so we get:

$F=4(\frac{400}{h})+5h$

which simplifies to

$F=\frac{1600}{h} +5h$

Now, this is the equation we need to minimize. In order to do so, we need to take the derivative of that and set it equal to zero, so we get:

$F=\frac{1600}{h}+5h\\ \\F=1600h^{-1}+5h\\ \\F'=-1600h^{-2}+5\\ \\F'=-\frac{1600}{h^{2}}+5$

And now we can set all this equal to zero, so we get:

$-\frac{1600}{h^{2}}+5=0$

Which can now be solved for h, so we get:

$-\frac{1600}{h^{2}}+5=0\\ \\-\frac{1600}{h^{2} }=-5\\ \\1600=5h^{2}\\ \\h^{2}=\frac{1600}{5}\\ \\h^{2}=320\\ \\h=\sqrt{320}=8\sqrt{5}m=17.89m$

That's where the first answer came from. We can now use this value to find the second answer together with the first equation:

$b=\frac{400}{h}\\ \\b=\frac{400}{8\sqrt{5}}\\ \\b=\frac{50}{\sqrt{5}}$

which simplifies to:

$h=\frac{50}{\sqrt{5}}*\frac{\sqrt{5} }{\sqrt{5} }\\ \\h=\frac{50\sqrt{5} }{5}\\ \\h=10\sqrt{5}m = 22.36m$

Aproximately and that's where te second answer came from.

So each  pen must measure:

22.36 m by 17.89 m

When multiplied you'll see that each pen will have an area of $400m^{2}$