#### Question

If a,b are real numbers such that a^2+b^2 = 5, find the maximum value of 3a+2b

Collected in the board: Quadratic function

Steven Zheng posted 4 months ago

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}

Steven Zheng posted 4 months ago

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}

Steven Zheng posted 4 months ago

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}

Steven Zheng posted 4 months ago

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