Question

A farmer wants to make a rectangular field with a total area of 800 m^2. It is surrounded by a fence. It is divided into 3 equal areas by fences. What is the shortest total length of fence with which this can be done?

Answer 1

calculus max min problem right?

Answer 2

\[3xy=800\\
y=\frac{800}{3x}\] and the total length is \[6x+2y\]

Answer 3

making your function
\[6x+\frac{1600}{3x}\] minimize that one

Answer 4

ok so now I would solve for x..?
or find the derivative then make it equal to 0?

Answer 5

Let w be the width of the field and l be the length of the field. We know \[wl=800 \]The total length of fence is going to be the perimeter (2w+2l), plus the 2 pieces of fencing in the middle, so 4w+2l (it doesn't matter if we say the middle pieces are w or l). So:
\[P=4w+2l\]We want to minimise P.
\[wl=800, w=\frac{800}{l}\]\[P=4w+2l=4\frac{800}{l}+2l\]\[P'=-3200l^{-2}+2l\]Setting P'=0 yields: \[0=-3200l^{-2}+2l\]\[3200l^{-2}=2l\]\[1600=l^3\]\[l=\sqrt[3]{1600} = 11.696 (3dp)\]Putting this back into our initial condition wl=800, gives us:
\[w*11.696=800\]\[w=68.399\]Hence solving for P give us:\[P=4w+2l=4(68.399)+2(11.696)=296.988\]

(I know P doesn't strictly define a perimeter here, I just thought it was the most appropriate word to use)

Answer 6

@satellite73, your total is 6x+4y is it not? You've left out the pieces in the middle.

Answer 7

wait but how would 296 be the shortest side?

Answer 8

That's the total fence length.

Answer 9

The sides are W=68.399 and L=11.696

Answer 10

but it's still coming up as incorrect

Answer 11

I messed up my derivative, sorry. I'm just working it again now.

Answer 12

We want to minimise \[P=4x+2y\]Subject to the constraint \[xy=800\]
\[y=\frac{800}{x}\]\[P=4x+2\frac{800}{x}\]\[P'=0=4-1600x^{-2}\]\[x^2=400\]\[x=20\]\[y=\frac{800}{x}=\frac{800}{20}=40\]
\[P=4x+2y=4*20+2*40=160\]