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a+b+c = 3,then find the greatest value of a^2b^3c^2.

a+b+c = 3,then find the greatest value of a^2b^3c^2.

Grade:12

2 Answers

Nishant Vora IIT Patna
askIITians Faculty 2467 Points
6 years ago
Hello Student, Please find the solution

Given a+b+c =3
Write this as
\frac{a}{2} + \frac{a}{2}+ \frac{b}{3}+ \frac{b}{3}+ \frac{b}{3}+ \frac{c}{2}+ \frac{c}{2} = 3

Now we will use AM >= GM

So
\frac{[\frac{a}{2} + \frac{a}{2}+ \frac{b}{3}+ \frac{b}{3}+ \frac{b}{3}+ \frac{c}{2}+ \frac{c}{2} ]}{7}} >= [\frac{a}{2}*\frac{a}{2}*\frac{b}{3}*\frac{b}{3}*\frac{b}{3}*\frac{c}{2}*\frac{c}{2}]^\frac{1}{7}

Therefore,
\frac{3}{7} >= [\frac{a}{2}*\frac{a}{2}*\frac{b}{3}*\frac{b}{3}*\frac{b}{3}*\frac{c}{2}*\frac{c}{2}]^\frac{1}{7}
[\frac{3}{7}]^7 >= [\frac{a}{2}*\frac{a}{2}*\frac{b}{3}*\frac{b}{3}*\frac{b}{3}*\frac{c}{2}*\frac{c}{2}]
[\frac{3}{7}]^7 >= \frac{a^2b^3c^2}{2^23^32^2}
[\frac{3}{7}]^7*2^23^32^2 >=a^2b^3c^2
So greatest value of a^2b^3c^2 is
\frac{3^{10}2^4}{7^7}
ankit singh
askIITians Faculty 614 Points
10 months ago

iven: a + 2b + 3c = 4

=> a = 4- 2b - 3c

To find : Minimum value of a² + b² + c² OR to find min value of (4–2b - 3c)² + b² + c² , here we have 2 variables related by the above shown expression. Let's suppose its minimum value is ‘m'

m = (4–2b-3c)² + b² + c² . . . . . . . . . . . (1)

The derivative of m wrt b = Polynomial can be differentiated by power rule (d/dx) x^n = n x^(n-1)

dm/ db = (-2) (2) ( 4–2b - 3c)^(2–1) + 2b=0 ( by quotient rule)

=> -16 + 8b +12c + 2b = 0

=> 10b + 12c = 16

=> 5b + 6c = 8 . . . . . . . . . (2)

Next, the derivative of m wrt c =

dm/dc = (-3) (2) ( 4–2b - 3c)^(2–1) + 2c =0

=> -24 + +12b +18c + 2c = 0

=> 12b + 20c = 24

=> 3b + 5c = 6 . . . . . . . . . . . .(3)

By solving eq (1) & eq(2)

We get b= 4/7 , c = 6/7

Since, ‘m’ is the minimum value of the expression

So, we substitute the values of b & c in eq (2) & (3), to get the value of m.

(4 - 2*4/7 - 3*6/7 )² + (4/7)² + (6/7)²

= ( 4- 8/7 - 18/7)² + 16/49 + 36/49

= 4/49 + 16/49 + 36/49

= 56/49

= 8/7

 

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