To determine the derivative of the area \( A \) with respect to \( x \), the product and chain rules are applied. Given \( A(x) = \frac{1}{2} \times x \times \sqrt{h^2 - x^2} \), the differentiation with respect to \( x \) yields: \[ \frac{dA}{dx} = \frac{1}{2} \times \left[ \sqrt{h^2 - x^2} + x \times \frac{d}{dx} \left( \sqrt{h^2 - x^2} \right) \right] \] Applying the chain rule to the second term: \[ \frac{d}{dx} \left( \sqrt{h^2 - x^2} \right) = \frac{-2x}{2\sqrt{h^2 - x^2}} = \frac{-x}{\sqrt{h^2 - x^2}} \] Substituting this back, we get: \[ \frac{dA}{dx} = \frac{1}{2} \times \left[ \sqrt{h^2 - x^2} - \frac{x^2}{\sqrt{h^2 - x^2}} \right] \] Simplifying the expression: \[ \frac{dA}{dx} = \frac{h^2 - 2x^2}{2\sqrt{h^2 - x^2}} \] To find the critical point, we set \( \frac{dA}{dx} = 0 \): \[ h^2 - 2x^2 = 0 \] \[ x^2 = \frac{h^2}{2} \] \[ x = \frac{h}{\sqrt{2}} \] Therefore, the critical point is located at \( x = \frac{h}{\sqrt{2}} \).