Concept:First Order Derivative:Lets say we have a function f which is continuous at the critical point, defined in an open interval I and f(c) = 0 (slope is 0 at c). Then we check the value of f'(x) at the point left to the curve and right to the curve and check the nature of f'(x), then we can say, that the given point will be:Local maxima: If f(x) changes sign from positive to negative as x increases via point c, then f(c) gives the maximum value of the function in that range.Local minima: If f(x) changes sign from negative to positive as x increases via point c, then f(c) gives the minimum value of the function in that range.Point of inflection: If the sign of f(x) doesnt change as x increases via c, and the point c is neither the maxima nor minima of the function, then the point c is called the point of inflection.Second-Order Derivative:If the second derivative of the function exists within the given range, then the given point will be:Local maxima: If f''(c) < 0Local minima: If f''(c) > 0Test fails: If f''(c) = 0Calculation:xy = 4225 ? y = \(\displaystyle \frac{4225}{x}\) ------(i)?Let S = x + y = x + \(\displaystyle \frac{4225}{x}\)?.Then \(\displaystyle \frac{ds}{dx}=\left(1-\frac{4225}{x^2}\right )\)? \(\displaystyle \frac{ds}{dx}=\left(\frac{x^2-4225}{x^2}\right) \)? \(\displaystyle \frac{ds}{dx}=0\) ?? \(\displaystyle \left(\frac{x^2-4225}{x^2}\right) =0\)? x2 = 4225 ? x = 65Also, from \(\displaystyle \frac{ds}{dx}=\left(1-\frac{4225}{x^2}\right) \), we get \(\displaystyle \frac{d^2s}{dx^2}=4225\left(\frac{2}{x^3}\right) =\frac{8450}{x^3}\)\(\displaystyle \frac{d^2s}{dx^2}\) is positive at x = 65, it has a minimum value at x = 65.Putting the value of x from equation (i), we get, ? y = \(\displaystyle \frac{4225}{x}=65\) ------(ii)?? S is minimum at x = 65 and minimum value of S = (65 + 65) = 130.