Find two positive numbers such that the sum of twice the first number and three times the second number is 240 and the product is a maximum

Answer 1

2x + 3y = 240

3y = 240 -2x

y = 80 - (2/3) x

x [ 80 - (2/3) x] = f(x)

80 x - (2/3) x^2 = f(x)

maximizing....

0 = 80 - (4/3) x = f'(x)

80*3/4 = x

x = 60

then y = 80 - (2/3)(60)

= 80 - 40

= 40

60*40 = 2400 is the max product