Answer:-24
explain how you know:
Since xy = 4, we have y=4/x. Substituting in (1) we get f(x) = 4x+(36/x) …….(2)
Differentiating we get f ' (x) = 4 - (36/x^2)…..(2)
For maxima or minima, f ‘ (x) = 0
=> 4 - (36/x^2) =0 => (4x^2 - 36)/x^2 = 0 which gives x= +/- 3
Differentiating (2), f ‘’ (x) = 72/x^3
When x= +3 , f ‘’ (x) is clearly +ve. Therefore f(x) at (2) gives minima and the minimum value is 24.
When x= -3, f ‘’ (x) is -ve. Therefore, f(x) has maxima at x= -3. The maximum value is -24.
Note: You may be surprised to observe that the minima is more than the maxima. This is due to the fact that the function is discontinuous at x= 0, the graph of f(x) comprising of two branches, one in the first quadrant giving the minima (= +24) and another in the third quadrant that gives maxima (= -24)
