Question

If \(a,b\gt 0\), then minimum value of \[ y=\frac{b^2}{a-x}+\frac{a^2}{x}, \qquad 0\lt x\lt a \] is

• A. \(4a\)
• B. \(4b\)
• C. \(2a\)
• D. \(2b\)

Hint:

For expressions involving sums of positive fractions, AM-GM inequality is often the quickest method: \[ u+v\geq 2\sqrt{uv}. \] Also remember that the quadratic \[ x(a-x) \] has maximum value \[ \frac{a^2}{4} \] at \[ x=\frac{a}{2}. \]

Answer 1

The Correct Option is B

Step 1: Write the given function.
Given, \[ y=\frac{b^2}{a-x}+\frac{a^2}{x}, \qquad 0\lt x\lt a \] We need to find its minimum value.

Step 2: Apply AM-GM inequality.
Using AM-GM, \[ u+v\geq 2\sqrt{uv} \] Take \[ u=\frac{b^2}{a-x} \] and \[ v=\frac{a^2}{x} \] Then, \[ y\geq 2\sqrt{ \frac{b^2}{a-x}\cdot \frac{a^2}{x} } \] \[ y\geq 2ab\sqrt{\frac{1}{x(a-x)}} \]

Step 3: Find maximum value of \(x(a-x)\).
Now, \[ x(a-x)=ax-x^2 \] This is a quadratic expression whose maximum value occurs at \[ x=\frac{a}{2} \] Substituting, \[ x(a-x)=\frac{a}{2}\left(a-\frac{a}{2}\right) \] \[ =\frac{a}{2}\cdot \frac{a}{2} \] \[ =\frac{a^2}{4} \] Hence, \[ \frac{1}{x(a-x)}\geq \frac{4}{a^2} \] Therefore, \[ \sqrt{\frac{1}{x(a-x)}}\geq \frac{2}{a} \]

Step 4: Obtain the minimum value.
Thus, \[ y\geq 2ab\cdot \frac{2}{a} \] \[ y\geq 4b \] Equality holds when \[ x=\frac{a}{2} \] Hence, the minimum value is \[ 4b \]

Step 5: Final conclusion.
Therefore, \[ \boxed{4b} \]