There are two ways to solve this: with calculus or without calculus. Both methods give the same result for x (the number of objects).
CALCULUS
Find df/dx=-6.4x+268.2. When df/dx is zero profit f(x) is maximum, so -6.4x+268.2=0, x=268.2/6.4=41.9 approx, so since x must be a whole number we take this to 42.
WITHOUT CALCULUS
Rewrite f(x)=-3.2(x²-268.2x/3.2+(134.1/3.2)²-(134.1/3.2)²)+257. We halve the coefficient of x and square it. This enables us to complete the square:f(x)=-3.2((x-134.1/3.2)²-(134.1/3.2)²)+257.
f(x)=-3.2(x-134.1/3.2)²+134.1²/3.2+257. Since the square of a number is always positive, the maximum value of f(x) is when x-134.1/3.2=0, because nothing is subtracted from the constant term, so x=134.1/3.2=41.9 approx, which rounds up to 42 objects.