Question
If a,b are real numbers such that a^2+b^2 = 5, find the maximum value of 3a+2b
Answer 1
If a,b are negative numbers such that 3a+2b becomes negative, then it is less than any positive numbers. Therefore, we are going to focus on the case a,b\geq 0
Given conditions
a^2+b^2 = 5
(1)
Construct the equation using 2a-3b
(3a+2b)^2+(2a-3b)^2
=9a^2+12ab+4b^2+4a^2-12ab+9b^2
=13(a^2+b^2)
Then we get the equation
(3a+2b)^2 = 13(a^2+b^2) -(2a-3b)^2
(2)
Substituting the given condition (1) gives
(3a+2b)^2 = 65 -(2a-3b)^2
(3)
It shows the (3a+2b)^2 has maximum value 65, if and only if 2a-3b=0, that is
a = \dfrac{3}{2}b
(4)
Substitute to (1) and solve for b
b = \sqrt{2}
and then
a = \dfrac{3}{2}\sqrt{2}
Therefore
When a = \dfrac{3}{2}\sqrt{2} , b=\sqrt{2} , 3a+2b has maximum value \sqrt{65}
Answer 2
Using Cauchy–Schwarz inequality
(x^2_1+y^2_1)(x^2_2+y^2_2)\geq (x_1x_2+y_1y_2)^2
Here
x_1=a,x_2 = b, x_2 = 3, y_2 = 2
Then
(a^2+b^2)(9+4)\geq ( 3a+2b)^2
Therefore
(3a+2b)^2\leq 13 (a^2+b^2) = 13\cdot 5 = 65
3a+2b\leq \sqrt{65}
The equation becomes an equality, if and only if x_1y_2 = x_2y_1, that is,
2a = 3b
The maximum value of 3a+2b is \sqrt{65}
Answer 3
Let
A = 3a+2b
then
a = \dfrac{A-2b}{3}
Substitute to given equation
a^2+b^2 = 5
A quadratic equation in terms of b is obtained
\Big( \dfrac{A-2b}{3} \Big) ^2+b^2 = 5
Expand and simplify the equation
13b^2-4Ab+A^2-45 = 0
In order for the quadratic equation to hold true, the discriminant must be greater than or equat to zero, that is
\Delta = (-4A)^2 - 4\cdot 13\cdot (A^2-45)\geq 0
then,
4(4A^2-13A^2+13\cdot 45 )\geq 0
A^2\leq 65
Taking square root on both sides gives
A \leq \sqrt{65}
We get the same result - the maximum value of 3a+2b is \sqrt{65}