Zachary R. answered 06/23/17
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General Mathematics and Sciences Tutor from K-Graduate
First draw a general form for your fence, then we will set the perimeter equal to 3480 yd.(accounting for the river) and finally optimize area as a function of a length somewhere(doesn't matter, it'll come out the same). If this comes with an image, use it. You can set the right sides to the river as a variable, L. the river and the side opposite the river will be W. Whether the dividing fence runs parallel or perpendicular to the river will determine its variable but I'll be assuming it's running right to the river, so it's length will be L As well. Thus your fencing is now 3"L"'s and 1 "W", which you can write as an expression 3L+W. This is your fencing, and you'll be using it all, waste not, want not. Thus you find
3L+W=3480 (1)
(I will use is and "=" interchangably). Now you have your hypothetical fence, it's nice. What is it's area? Area is the product of length and width so A=L•W (2)
All well and good but you cannot derlive th area with two variables. So you need to write one variable in terms of the other. Solving for W gives W=3480-3L (3)
from (1). Substituting W from (3) into (2) gives you
A=L•(3480-3L) (4)
Now there's two methods to solve this depending on if you know calculus. You can calculus through, derive and find your zero and test if this is a maximum, or use algebra to find the vertex since (4) is quadratic. I'll show both
Calculus
deriving A from (4) gives
dA=3480-6L (5)
Since you only need the extrema of this dA=0 and L=3480/6
Algebra
Multiplying out (4) gives
A=-3L^2+3480L
The vertex is the highest point and
-b/(2a) so L is 3480/6 and W is 3480/2