Find two positive numbers that satisfies the given requirements. The product is 147 and the sum of the first number plus three times the second number is a minimum.

Question

Find two positive numbers that satisfies the given requirements. The product is 147 and the sum of the first number plus three times the second number is a minimum.

Answer 1

This would be a harder problem if a and b were allowed to be "both negative" or "both positive." When negative the sum reaches a local maximum at b=-7 so all other b<0 have sums less.

But sticking to positive a and b. S(b)=(147/b)+3b

S'(b)=-147b^-2+3,,,,,critical point for b>0 is b=7

S''(b)=294b^-3 which is positive for b=7 meaning b=7 is a local minimum.

Answer 2

Two numbers: a and b

Their product is equal to 147 that is a * b = 147

Sum of the first and 3 times the second: a + 3b should be minimal

From the first equation we have a = 147/b and a+3b = 147/b+3b, So we need to find minimal value of the last expression. The minimal value is reached at b=-7 or b=7 (can be checked by differentiation). Then find the value of a.

Hope this helps