The sum of three positive numbers is 60. The first plus twice the second plus three times the third
add up to 120. Find the numbers which maximized the product all the three numbers.
x+y+z = 60
x+2y+3z = 120
x-z = 0
so x=z
and we have
f(x) = x^2(60-2x)
f'(x) = 120x - 6x^2 = 6x(20-x)
so the numbers are 20,20,20
as usual, maximum volume of a prism (for a given surface area) is when it is a cube.
why does x-z became equal to zero?
To maximize the product of three numbers, we need to find the numbers that would give the largest possible result.
Let's represent the three numbers as x, y, and z.
From the given information, we can form the following equations:
Equation 1: x + y + z = 60
Equation 2: x + 2y + 3z = 120
To solve this system of equations, we can use substitution or elimination method. Let's use the substitution method.
1. Solve Equation 1 for x: x = 60 - y - z
2. Substitute this value of x into Equation 2:
60 - y - z + 2y + 3z = 120
3. Simplify the equation:
60 + y + 2z = 120
y + 2z = 60
4. Solve Equation 3 for y: y = 60 - 2z
5. Substitute this value of y into Equation 1:
x + (60 - 2z) + z = 60
x - z = 0
x = z
Now we have expressions for x, y, and z in terms of z. Let's substitute x = z and y = 60 - 2z back into Equation 1 to find the values of x, y, and z.
x + y + z = 60
z + (60 - 2z) + z = 60
60 - z = 60
z = 0
Now, we can find the values of x and y using the derived expressions:
x = z = 0
y = 60 - 2z = 60 - 2(0) = 60
Therefore, the numbers that maximize the product are x = 0, y = 60, and z = 0.