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?

1 Answer
VNVDVI
Apr 6, 2018

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

Explanation:

Differentiate, with respect to x.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.

Related questions
Impact of this question
2076 views around the world
You can reuse this answer
Creative Commons License