How to prove x(x - c) is maximized when x = c/2?

Question

Does this have anything to do with the quadratic formula? What's a good way to prove this?

Answer 1

It isn't maximized when x = c/2. It's minimized there, though. A little bit of calculus will give us the proof:

1) Local extrema of a function f(x) can only occur at points where the first derivative f'(x) = 0. So let f(x) = x(x-c) = x^2 - cx. Then f'(x) = 2x - c. Setting f'(x) = 0 gives you 2x - c = 0. Bring c to the right-hand side to get that 2x = c, or c = x/2. So this is the only point on the whole real line where f(x) can have a local maximum or minimum.
2) To determine if f has a local maximum or a minimum x = c/2, we use the 2nd derivative test. f''(x) = 2, which is strictly positive. By the 2nd derivative test, then, f has a local minimum at c/2.

*EDIT: It turns out to be the global minimum, and it looks like a few people who got here before me have given good reasons why.*

Answer 2

To find maxima and minima you can find the values for which the derivative is equal to zero.

Which here means: 2x - c = 0 or x = c/2.

To make sure this is a maximum and not a minimum you calculate the value for x = c/2 and another value (usually x=0 is an easy one.
for x = c/2 you get -c^2/2 fo x=0 you get 0. In fact -c^2/2 < 0 so it is a minimum.

Answer 3

For maximum/minimum problems, we need to find the first and second derivative of the function
fn(x) is x(x-c)
call the function y
then y= x^2-x(c)
and 1st derivative with respect to x is 2(x) -c
the 2nd derivative is 2, which is positive

therefore a minimum occurs when the first derivative equals 0.
so a min at 2(x) =c =0
or x = c/2.

Answer 4

x^2 - cx

Find derivative and set to 0

2x - c = 0

2x = c

x = c/2

It's minimized by the way.

_______________________________

Since you don't seem to know calculus I'll each you a bit.

A derivative is the instantaneous rate of change at a point.

A line will always have a constant rate of because it will change according to a constant slope.

Other functions such as x^2 and x^3, or etc change according to x.

Power rule:

Derivative of x^n = n(x^(n-1))

Deivative of a sum = derivative of the terms added toe ach other.

Derivative of cx = c

The derivative of a maximum or minimum point will always be 0, because they can't go any higher or can anywhere lower.

If you asked why, that would be simple. BECAUSE IT'S MAXED OUT AND MINIMIZED OUT.

This is the proof of why parabola have maximum at a certain point.

y = ax^2 + bx + c

Differentiate both sides and derivative to 0

0 = 2ax + b

(y = c is always a horizontal line with no slope, so their derivative is 0)

-b = 2ax

x = -b/2a

Plug that into y = ax^2 + bx + c and you'll get your maximum or minimum y value.

Answer 5

When expanded, the expression is x^2 - cx, which is a quadratic equation.

A quadratic equation's vertex (aka minimum if the coefficient of x^2 is positive, or maximum if the coefficient of x^2 is negative) is given by x= -b/(2a), where a is the coefficient of x^2 and the coefficient of x.

so x(x-c) does not have a maximum. Rather, it has a minimum at x = -(-c)/2 = c/2.

Answer 6

f(x) = x(x - c)
is a quadratic function.
that is a parabola.
it has two zeros : x = 0 and x = c
in the middle of the zeros it is the maximum ( or minimum ) thus c/2.

since the sign before the x^2 is positive , c/2 is a minimum.

Answer 7

(a million) and (2) questions are direct utility of discriminant of quadratic equation./////////// For (3) question A homogenious equation of the type ax^2+2hxy+via^2+2gx+2fy+c to be resolved because of the fact the fabricated from 2 components linear in x and y it may fulfill abc + 2fgh -af^2 - bg^2 - ch^2=0

Answer 8

y=x(x-c)
x^2-cx
dydx=2x-c=0, x=c/2
d^2y/dx^2=2=+ve
so x=c/2 gives the minimum value of y.