Question

The max value of the quadratic function
fG) = -3x*x+12x5
over the interval (3) is​

Answer 1

Answer:

ANSWER

Given, f(x)=2x

3

−9ax

2

+12a

2

x+1

Thus f

′

(x)=6x

2

−18ax+12a

2

and f

′′

(x)=12x−18a

For max/ min, 6x

2

−18ax+12a

2

=0

⇒x

2

−3ax+2a

2

=0

⇒(x−a)(x−2a)=0

⇒x=a or x=2a

Now, f

′′

(a)=12a−18a=−6a<0

and f

′′

(2a)=24a−18a=6a>0

Therefore, f(x) is maximum at x=a and minimum at x=2a.

⇒p=a and q=2a

Given that, p

2

=q⇒a

2

=2a

⇒a=2 (a>0).