Question 828225
Let x = the positive number. Then "one-fourth of its square" translates into:
{{{(1/4)x^2}}}.<br>
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-(1/4)x^2}}}<br>
Let y = the difference. So:
{{{y = x-(1/4)x^2}}}
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.<br>
When a quadratic is in standard form, {{{y = ax^2+bx+c}}}, the x-coordinate of the will be {{{(-b)/2a}}}. So we will first put our equation in standard form:
{{{y = -(1/4)x^2+x}}}
and then the x-coordinate of the vertex is:
{{{(-1)/2(-1/4)}}}
which simplifies as follows:
{{{(-1)/(-1/2)}}}
{{{((-1)/(-1/2))(2/2)}}}
{{{(-2)/(-1)}}}
{{{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 = -(1/4)x^2+x}}}. The resulting y value will be this maximum difference.)