The product is 768 and the sum of the first plus three times the second is a minimum.

Question

Find two positive numbers that satisfy the given requirements.

Answer 1

FS=768

F+3S = minimum

F=48

S=16

48x16=768

48+3(16) = 96=minimum

f(F,S) = F+3S

f(S) = 768/S + 3S

take the dervative

set it equal to zero

f'(S)= 3 -768/S2=0

multiply by S^2

3S^2 -768 =0

solve for S

3S^2 = 768

S^2 = 768/3 = 256

S =sqr256 = 16= Second Number

F=768/16=48=First Number

Answer 2

The two numbers are x and 796 / x

The sum is 3x + (796 / x), or (3x² + 796) / x.

f(x) = (3x² + 796) / x

Can you determine f'(x) and then minimum? (Division Rule)