Question

Find the maximum value of y= -8x^2 + 80x -196

Answer 1

Maximum value of

`f(x) = - 8x^2 + 80x - 196`

is

`\boxed{f(x) = 4}`

Explanation:

The maximum or minimum value of a function f(x) occurs when the first derivative equals zero

The first derivative of

`f(x) = - 8x^2 + 80x - 196`

`f'(x) = (d)/(dx)\left(-8x^2+80x-196\right)`

`=-(d)/(dx)\left(8x^2\right)+(d)/(dx)\left(80x\right)-(d)/(dx)\left(196\right)`

`(d)/(dx)\left(8x^2\right) = 16x\\\\=- (d)/(dx)\left(8x^2\right) = -16x\\\\\\(d)/(dx)\left(80x\right)=80\\\\(d)/(dx)\left(196\right)=0`

`(d)/(dx)\left(-8x^2+80x-196\right) \\\\\\ =-16x + 80 + 0\\\\= -16x + 80\\\\`

Set this expression equal to 0 and solve for x to find the x value which maximizes or minimizes the function

`-16x + 80 = 0\\\\\implies -16x = -80\\\\x = -80/-16\\\\x = 5\\\\`

Plug this value of x into the original function to get the maximum or minimum value.

`f(5) = -8(5^2) + 80(5) - 196\\\\= -8 (25) +400 -196\\\\= -200 + 400 - 196\\\\= 4\\\\`

So the maximum value of
`\boxed{f(x) = 4}` ANS

and occurs at
`x = 5`

(Strictly speaking to find if this is a maximum or minimum, we have to find the second derivative and see if it is negative or positive. If negative, it is a maximum, if positive it is a minimum

But since the question does not ask for this, I am skipping it. In case you are interested, the second derivative of f(x) is the derivative of f'(x) which will work out to -16 so it is a maximum)

Answer 2

(5,4)

Explanation:

first, if you have a calculator you can put the equation into y = and just find the maximum point there

to find the x value of the maximum point you do -B/2A

-80/2(8) = 5 so the x value would be 5

then to find y just plug in 5 for x and you should get 4