# Thread: Finding Value of x at which y is a maximum

1. ## Finding Value of x at which y is a maximum

y = -x^2 + 20x

Finding Value of x at which y is a maximum. What is the maximum value of Y.

This is obviously so easy but I need help because Im stupid and suffering from a severe mind blank.....In this circumstance will someone just point me in the right direction telling me what i need to do without doing it...obviously involves differentiation and possibly the second derivative

2. Originally Posted by rooney
y = -x^2 + 20x

Finding Value of x at which y is a maximum. What is the maximum value of Y.

This is obviously so easy but I need help because Im stupid and suffering from a severe mind blank.....In this circumstance will someone just point me in the right direction telling me what i need to do without doing it...obviously involves differentiation and possibly the second derivative
first of all, you should have a mental picture of the graph of y (a parabola with vertex up)

using calculus ...

find y' , set y' = 0 to find the x-value of the vertex (the maximum).

the maximum value is the value of y at that value of x.

using algebra ...

x-value of the vertex = -b/(2a)

using calculus

3. Take the derivative and solve for x. Which would be your relative max or relative min. Then taking the 2nd derivative and if it is f''>0 you have a local minimum. if f''<0 you have a local maximum and if f''=0 then you have a possible inflection point.

4. ok

so

-x^2+20x = 0
becomes -2x + 20 = 0
therefore the value of x = 10...correct?
then are you saying i need to sub into orig function which leaves y=300?

5. Originally Posted by rooney
ok

so

-x^2+20x = 0
becomes -2x + 20 = 0
therefore the value of x = 10...correct?
then are you saying i need to sub into orig function which leaves y=300?
its 100. -(10)^2 = -100 :P ur mind seems fried, take a break homie.

6. Originally Posted by rooney
ok

so

-x^2+20x = 0 you do not set the original function equal to 0, just the derivative
becomes -2x + 20 = 0
therefore the value of x = 10...correct?
then are you saying i need to sub into orig function which leaves y=300? that's right (the subbing in part), not the y=300
now, answer this ... why do you have to set the derivative equal to 0?

7. Originally Posted by jakncoke
its 100. -(10)^2 = -100 :P ur mind seems fried, take a break homie.
-2(10)+20 = 0

therefore = 10.

Sub 10 into original equation 10^2 + 20(10)= 300

hmmm.....i dont see how thats = 100?

8. y = -x^2 + 20x

y = -(10^2) + 20(10)

y = -100 + 200 = 100

9. -10^2 = 100!

10. no ... $\displaystyle -x^2 = -1 \cdot x^2$

11. -10^2 = -10 * -10 = 100

12. Originally Posted by rooney
-10^2 = -10 * -10 = 100
then what sense would it make to have a -x^2? instead of x^2 ? $\displaystyle -x^2 = -1 * x * x ....$ $\displaystyle -1 * 10 * 10 = -100$ ....

13. Originally Posted by rooney
-10^2 = -10 * -10 = 100
you are confusing $\displaystyle -x^2$ with $\displaystyle (-x)^2$ ... the two expressions are different. remember the order of operations?

14. You don't need calculus to find the maximum value of a quadratic function: complete the square.

$\displaystyle y= -x^2+ 20x= -(x^2- 20x)= -(x^2- 20x+ 100- 100)= -(x- 10)^2+ 100$.

If x= 10, x- 10= 0 so y= 100. If x is any other value, $\displaystyle (x- 10)^2$ is positive so $\displaystyle -(x- 10)^2$ is negative and $\displaystyle -(x-10)^2+ 100$ is less than 100.