With respect to optimization problems

[kimmy_koo51]

A farmer has 600m of fence he wants to enclose a rectangular field beside the river on his property. Find the dimensions of the field so that a maximum area is enclosed. (Fencing is required for only three of the sides).

So far...

Let:
P be perimeter
l be length
w be width
The river is along the length?
If so...
l + 2w = P
600m = P
600 = l +2w
l = 2w - 600
Sub this back into the equation for perimeter, expand, find the derivative, and then set it equal to zero?

 

[galactus]

Yes and no. Sub into the area formula. That's what you're minimizing.

Watch, you have a sign error at the bottom.

Perimeter: \(\displaystyle \L\\2x+y=600\).......[1]

Area: \(\displaystyle xy\)...........[2]

Solve [1] for y and sub into [2]:

\(\displaystyle A=x(600-2x)\)

Now, do that differential thing.

 

[kimmy_koo51]

Excellent, thank you...that was the easitest problem out of my homework, and the only one I didn't get. Haha.