1. ## Differentiation Application Problem

I think I have this one right but I'm not sure. Here's what I have

The problem states: A farmer plans to fence in a rectangular pasture containing three adjacent pens of equal size, lined up end-to-end. The total length of fencing needed to enclose the three pens is 1200 yards. Find the maximum total area that all three pens can enclose.

So I set up the equation as follows

$A=3xy$

$6x+4y=1200$

$y=\frac{-6x+1200}{4}$

$A=3x(\frac{-6x+1200}{4})=\frac{18x^2+3600x}{4}$

Apply Quotient Rule
$A'=\frac{-144x+14400}{16}=-9x+90$

$x=100$

so

$y=150$

$A=3(100)(150)$

$A=45000$

What did I do wrong?

Thanks

2. Originally Posted by beanus
I think I have this one right but I'm not sure. Here's what I have

The problem states: A farmer plans to fence in a rectangular pasture containing three adjacent pens of equal size, lined up end-to-end. The total length of fencing needed to enclose the three pens is 1200 yards. Find the maximum total area that all three pens can enclose.

So I set up the equation as follows

$A=3xy$

$6x+4y=1200$

$y=\frac{-6x+1200}{4}$

$A=3x(\frac{-6x+1200}{4})=\frac{18x^2+3600x}{4}$

Apply Quotient Rule
$A'=\frac{-144x+14400}{16}=-9x+90$
First you apparently lost a minus sign between the second to last and last line. Also, notice that 16 is a constant, so you don't use the quotient rule. So....
$\displaystyle A = \frac{-18x^2 + 3600x}{4}$

$\displaystyle A' = \frac{-36x + 3600}{4} = -9x + 900$

Now, you still come out with x = 100, so there is damage, but the end result is the same. You still get y = 150, giving A = 45000. Why do you think something is wrong?

-Dan

3. Yeah the missing sign was a typo, sorry. I'm wondering if it is right or not because the online math program I'm using says I'm wrong, and I'm pretty sure I'm right. Thanks Dan!