Two positive numbers x, y, such that x + 2y = 160.
The values of x and y that maximize xy.
The product will be maximized when its derivative is zero. We can write the product in terms of a single variable to make solution easier.
... y = (160 -x)/2 . . . . . solve the given relation for y
... f(x) = x·y = x(160 -x)/2 . . . . . substitute the expression for y into the product
... f'(x) = 80 -x = 0 . . . . . . . . . . find the derivative of f(x) and set it to zero
... x = 80 . . . . . . . . . . . . . . . . . add 80
... y = (160 -80)/2 = 40 . . . . . . find the value of y
The first number is 80.
The second number is 40.
