I know the answer is 5000 and that the equation to figure out the answer is in quadratic form. I am just not sure on how to get to the answer, so it would be great if someone could show the steps on how to get to the answer. It would be greatly appreciated.
Let x be the length of the side parallel to the barn.
The other two side lengths are therefore 100 - x/2
Write the equatuion for Area as a function of x, and set the derivative dA/dx = 0. That will let you solve for x. It will turn out to be 100 yards. The area will then be 50*100 = 5000 sq yds
I understand how you got the answer, but I need to start with 2W+L=200 and somehow turn that into a quadratic equation(ax^2+bx+c) to obtain the answer.
Let x = L
Then W = (200 - x)/2 = 100 - (x/2)
That is exactly what I did. It does not matter whether I call the length x or L. The algebra is the same.
The area is then A(x) = 100 x - x^2/2
If they don't want you to use calculus, then complete the square to find the x that gives maximum Area.
2 A(x) = 200 x -x^2
= -(x-100)^2 + 10,000
That clearly has a maximum whan x = 100, and that value corresponds to 2A = 10,000. or A = 5,000
1=(X-100)/((X+100)/2)) / ((2-1)/((2+3)/2)
Solve for X
I know X =200, but don't understand how to reach that anwser.
is this really algebra, im doing this now in freshman year algebra 2