# 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. - Mathematics

If the function f(x)=2x39mx2+12m2x+1, where m>0 attains its maximum and minimum at p and q respectively such that p2=q, then find the value of m.

#### Solution

We have
f(x)=2x39mx2+12m2x+1

f'(x)=6x218mx+12m2
Also, f''(x)=12x18m
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

6p218mp+12m2=0

p23mp+2m2=0

(p2m)(pm)=0

p2m =0 or pm=0

p=2m or p=m

Similarly,
f '(q) = 0
q=2m or q=m
Now, consider the following cases:
Case I:
If p = 2m and q = 2m, then

p2=q

4m2=2m

2m2m=0

m(2m1)=0

m=1/2      (m>0)
But, this gives p = 1 as the point of minima, which is not true.

Case II:
If p = 2m and q = m, then
p2=q

4m2=m

4m2m=0

m(4m1)=0

m=1/4      (m>0)
But, this gives p = 12 as the point of minima, which is not true.

Case III:
If pm and q = 2m, then
p2=q

m2=2m

m22m=0

m(m2)=0

m=2      (m>0)
For this case, p = 2 and q = 4 are the points of maxima and minima, respectively.

Case IV:
If pm and qm, then
p2=q

m2=m

m2m=0

m(m1)=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.

Concept: Simple Problems on Applications of Derivatives
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