Question

Maximum of 36x-x^3

Answer 1

find the 1st derivative

\[\frac{dy}{dx} =36 - 3x^2\]

then solve for x to find the stationary points

\[36 - 3x^2 = 0...... 2(\sqrt{12} -x)(\sqrt{12} + x) = 0\]

stationary points exist at

\[x = \pm \sqrt{12}\]

find the 2nd derivative

\[\frac{d^2y}{dx^2} = - 6x\]

test the stationary points in the 2nd derivative

\[\frac{d^2y}{dx^2} = -6(-\sqrt{12} .....>0....\]

minimum

\[\frac{d^2y}{dx^2} = -6(\sqrt{12})....<0 ...\]

maximum exists at \[x = \sqrt{12} \]

substitute it into the original equation to find the maximum value of y

\[y = 36\times \sqrt{12} - (\sqrt{12})^3\]

Answer 2

if you dont know calc find the roots and the max should be in the middle of the 2 roots.

so say your roots are 0 and 3 then max should be at x=1 :)

Answer 3

the problem has 3 roots so which 2 do you choose