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=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.

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