Question

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

Answer 1

Correct Answer: 20

To find the minimum possible value of the function \( f(x) = \max \left\{5x,\ 52 - 2x^2 \right\} \), we first need to determine the point where the two expressions \( 5x \) and \( 52 - 2x^2 \) are equal.

Step 1: Solve \( 5x = 52 - 2x^2 \) 

Rearranging the equation: \( 2x^2 + 5x - 52 = 0 \)

Step 2: Apply the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2 \), \( b = 5 \), \( c = -52 \)

Step 3: Compute the discriminant: \( b^2 - 4ac = 5^2 - 4 \cdot 2 \cdot (-52) = 25 + 416 = 441 \)

Step 4: Since 441 is a perfect square, \( x = \frac{-5 \pm \sqrt{441}}{4} = \frac{-5 \pm 21}{4} \)

Step 5: Solve for \( x \): \( x = \frac{16}{4} = 4 \) or \( x = \frac{-26}{4} = -6.5 \)

Since we are interested in positive values of \( x \), we take \( x = 4 \).

Step 6: Evaluate both expressions at \( x = 4 \):

• \( 5x = 5 \cdot 4 = 20 \)
• \( 52 - 2x^2 = 52 - 2 \cdot 16 = 20 \)

Thus, at \( x = 4 \), both expressions equal 20. Therefore, the minimum possible value of \( f(x) \) is:

\( \boxed{20} \)