Question

Let f(x) = min{2x2 ,52−5x}, where x is any positive real number. Then the maximum possible value of f(x) is

Answer 1

Correct Answer: 32

We are given: \[ f(x) = \min\left( 2x^2, \ 52 - 5x \right) \] To find the maximum possible value of \( f(x) \), we first determine where: \[ 2x^2 = 52 - 5x \] 

Step 1: Rearrange the equation

\[ 2x^2 + 5x - 52 = 0 \]

Step 2: Apply the quadratic formula

Here, \( a = 2 \), \( b = 5 \), \( c = -52 \): \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] \[ x = \frac{-5 \pm \sqrt{25 - 4(2)(-52)}}{4} \] \[ x = \frac{-5 \pm \sqrt{25 + 416}}{4} \] \[ x = \frac{-5 \pm \sqrt{441}}{4} \]

Step 3: Solve

\[ x = \frac{-5 \pm 21}{4} \] This gives: \[ x = \frac{16}{4} = 4 \quad \text{or} \quad x = \frac{-26}{4} = -\frac{13}{2} \] Since \( x \) must be positive, we take \( x = 4 \).

Step 4: Evaluate \( f(4) \)

\[ 2x^2 = 2(16) = 32 \] \[ 52 - 5x = 52 - 20 = 32 \] Both expressions equal 32 at \( x = 4 \).

✅ Therefore, the maximum possible value of \( f(x) \) is 32.