Question

Find two positive numbers whose product is 400 and such that the sum of twice the first and three times the second is a minimum. Optimization Problem

Answer 1

equations are
x y = 400
minimize 2 x + 3 y

convert problem to single variable using constraint
y = 400/x

rewrite goal
minimize f(x) = 2 x + 3 (400/x)

set derivative =0
f'(x) = 2 -1200 x^-2 [two minus 1200 divided by square of x]
2 x^2 = 1200
x = sqrt(600) = 24.5

y = 400/x= 16.3

min f = 2(24.5) + 3(16.3) = 97.9

to be sure this is a minimum, take second derivative and show that it is
positive over the domain.

f"(x) = 2400 x^-3 positive for x>0 Does check.

That's the method. Try it yourself to be sure I have not calculated anything wrong.

Answer 2

Thanks