If x and y are positive real numbers such that xy = 64, then the leas…

The correct answer is Option (4) → 32

Since \(y = \frac{64}{x}\), the sum depends on x alone:

    \( \displaystyle S(x) = x + \frac{256}{x}, \quad x \gt 0 \)

Setting the derivative equal to zero:

    \( \displaystyle S'(x) = 1 - \frac{256}{x^{2}} = 0 \)

    \( x^{2} = 256 \), so x = 16 and y = 4

Since \( \displaystyle S''(x) = \frac{512}{x^{3}} \gt 0 \), this is a minimum:

    \( S(16) = 16 + 16 = 32 \)

Note that x = y = 8 gives 8 + 32 = 40, which is not the least value: the minimum of x + 4y is not at equal numbers.