Problema Solution
The sum of three positive numbers is 30. The first plus twice the second plus three times the third add up to 60. Delver the numbers so that the product of all three is as large as possible.
Answer provided by our tutors
let
x = the first number
y = the second number
z = the third number
The sum of three positive numbers is 30:
x + y + z = 30
The first plus twice the second plus three times the third add up to 60:
x + 2y + 3z = 60
using the system
x + y + z = 30
x + 2y + 3z = 60
we solve for y and z and find:
y = -2x + 30
z = x
click here to see the step by step solution of the system of equations:
xyz = x(-2x + 30)x = -2x^3 + 30x^2
we also know that y > 0 that is -2x + 30 > 0
by solving the inequality we find:
x < 15
click here to see the step by step solution of the inequality:
thus the maximum of -2x^3 + 30x^2 is for x = 14 and is equal to -2*14^3 + 30*14^2 = 392
y = -2*14 + 30
y = 2
z = 14
the numbers are 14, 2 and 14.