If the function f(x)=2x^{3}−9mx^{2}+12m^{2}x+1, where m>0 attains its maximum and minimum at *p* and *q* respectively such that p^{2}=q, then find the value of *m*.

#### Solution

We have

f(x)=2x^{3}−9mx^{2}+12m^{2}x+1

⇒f'(x)=6x^{2}−18mx+12m^{2}

Also, f''(x)=12x−18m

Since, *f*(*x*) attains its maximum and minimum values at *x* = *p* and *x* = *q*, respectively, so *f *'(*p*) = 0 and

*f *'(*q*) = 0*f *'(*p*) = 0

⇒6p^{2}−18mp+12m^{2}=0

⇒p^{2}−3mp+2m^{2}=0

⇒(p−2m)(p−m)=0

⇒p−2m =0 or p−m=0

⇒p=2m or p=m

Similarly,*f *'(*q*) = 0

⇒q=2m or q=m

Now, consider the following cases:

Case I:

If* p* = 2*m* and *q* = 2*m*, then

p^{2}=q

⇒4m^{2}=2m

⇒2m^{2}−m=0

⇒m(2m−1)=0

∴m=1/2 (m>0)

But, this gives *p* = 1 as the point of minima, which is not true.

Case II:

If* p* = 2*m* and *q* = *m*, then

p^{2}=q

⇒4m^{2}=m

⇒4m^{2}−m=0

⇒m(4m−1)=0

∴m=1/4 (m>0)

But, this gives *p* = 12 as the point of minima, which is not true.

Case III:

If* p* = *m* and *q* = 2*m*, then

p^{2}=q

⇒m^{2}=2m

⇒m^{2}−2m=0

⇒m(m−2)=0

∴m=2 (m>0)

For this case, *p = *2 and *q* = 4 are the points of maxima and minima, respectively.

Case IV:

If* p* = *m* and *q* = *m*, then

p^{2}=q

⇒m^{2}=m

⇒m^{2}−m=0

⇒m(m−1)=0

∴m=1 (m>0)

But, this gives *q* = 1 as the point of maxima, which is not true.

Hence, the value of m is 2.