Question

Area is A=30x-x^2 x can vary a)Find value of x at which A is stationary (I have found it's 15) b)find this statinary value of A and determine whether it's maximum or minimum value?

Answer 1

The stationary value `x=15` is a maximum value.

Explanation:

Differentiate, with respect to `x.`

`A'=30-2x`

As you've determined, the stationary value (value for which `A'=0`) is `15.`

`30-2x=0`

`2x=30, x=15`

Now, take the second derivative.

`A''=-2`

The Second Derivative Test tells us that if the second derivative is positive at a stationary point, that stationary point is a minimum; if the second derivative is negative at a stationary point, that stationary point is a maximum.

Well, `A''=-2<0`, always, regardless of the value of `x.` This tells us that ANY stationary value will be a maximum value.