Two similar triangular pastures meet at a
vertex, The dimensions are glven in yards,
How much fencing would be needed to
enclose the larger pasture?
11
7
5
13
x
у
A 28.2 yards
42.7 yards
47 yards
59.8 yards

Note that this is still a rectangle, and the length of the fencing is still 600 ft.

So the sum of the sides NOT along the river x + x + y = 600, and the area equals xy.

This makes the two equations: 2x + y = 600, and A = xy.

To find the largest area, we need to find A as a function of x or y. I suggest solving the first equation for y and replacing that in the second equation.

y = 600 - 2x. and A(x) = x(600-2x)

We now need to maximize A(x) = 600x - 2x2.

Remember, if x = -b/(2a), we find the x value of the vertex, the y value can be found by substitution.

So, since a = -2, and b = 600, x = -600/(-4) = 150 ft. If x = 150, y = 600 - 2(150) = 300.

So, the dimensions are 150 x 300 and the maximum area = 300(150) = 45,000 ft2

I hope this helps. By the way, there are many variations of this, and they are all similar. For example, you might want to make several pens with two lengths parallel and have three parallel withs inside.