CONCEPT: Second Derivative Test: Let f be a function defined on an interval I. - Calculate f′(x) - Solve f′(x)=0 and find the roots of f(x)=0. Suppose x=c is the root of f′(x)=0 - Calculate f′′(x) and put x=cto get the value of f′′(c). - If f " (c)<0 then x=c is a point of local maxima. -If f′′(c)>0 then x=c is a point of local minima. - If f"(c)=0 then we need to use the first derivative test. CALCULATION: Given: x+y=20 and P=xy ⇒P=x⋅(20−x)=20x−x2 ⇒P(x)=20x−x2 ⇒P′(x)=20−2x If P′(x)=0⇒20−2x=0⇒x=10 By substituting x=10 in the equation x+y=20 we get, y=10 ⇒P′′(x)=−2 ⇒Pn(x=10)=−2<0 So, x=10 is the point of maxima So, the maximum value of P=xy=10×10=100 Hence, option 1 is correct.