SOLUTION: What positive number exceeds one-fourth of its square by a maximum amount?

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Question 828225: What positive number exceeds one-fourth of its square by a maximum amount?
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
Let x = the positive number. Then "one-fourth of its square" translates into:
%281%2F4%29x%5E2.

The amount one number is more than another number is also called the difference between the numbers. Differences are found by subtracting. So the amount that our number, x, exceeds one-fourth its square is:
x-%281%2F4%29x%5E2

Let y = the difference. So:
y+=+x-%281%2F4%29x%5E2
The problem asks us to find the x that creates the largest difference (aka "y"). The graph of our equation is a parabola which opens downward (because of the "-" in front of the squared term. So the maximum difference, y, will at the vertex of this parabola. So our task becomes: Find the x-coordinate of the vertex (because it will create the largest possible y/difference.

When a quadratic is in standard form, y+=+ax%5E2%2Bbx%2Bc, the x-coordinate of the will be %28-b%29%2F2a. So we will first put our equation in standard form:
y+=+-%281%2F4%29x%5E2%2Bx
and then the x-coordinate of the vertex is:
%28-1%29%2F2%28-1%2F4%29
which simplifies as follows:
%28-1%29%2F%28-1%2F2%29
%28%28-1%29%2F%28-1%2F2%29%29%282%2F2%29
%28-2%29%2F%28-1%29
2
So 2 is the positive number which exceeds one-fourth of its square by the maximum amount. (If you want to find this maximum amount, just put a 2 in for the x in our equation: y+=+-%281%2F4%29x%5E2%2Bx. The resulting y value will be this maximum difference.)