The sum of three positive number is 30. The first plus twice the second plus three times the third add up to 60. Select the numbers so that the product of all three is as large as possilbe.
just write down the math for the words:
x+y+z = 30
x+2y+3z = 60
Think of the product as the volume of a rectangular solid. The solid has maximum volume when it is a cube. SO, we want x,y,z to be nearly equal in value.
1st guess: x=y=z=10
That also fits the 2nd equation.
So, xyz has a maximum value of 1000.
To check that, suppose we adjust x and y by an amount a. Then we have
(10-a)(10+a)(10)
= 10(100-a^2)
= 1000 - 10a^2
which is less than 1000
Well, this sounds like a math problem, but let's clown around with it!
To find the numbers that will give us the largest product, we need to play a little guessing game. Let's call the three numbers a, b, and c.
From the given information, we know that a + b + c = 30 and a + 2b + 3c = 60. Let's rearrange the second equation to solve for a in terms of b and c.
a = 60 - 2b - 3c
Now, let's substitute this value of a into the first equation:
(60 - 2b - 3c) + b + c = 30
Simplifying this equation, we get:
-2b - 2c = -30
Dividing both sides by 2, we have:
-b - c = -15
Well, it seems like the equations are getting a bit complicated. But don't worry, we're clowns, and we can simplify things!
Let's choose a value for c and then solve for b:
If we let c = 1, then -b - 1 = -15
Simplifying, we get:
b = 14
Now, we can substitute the values of b and c into the equation for a:
a = 60 - 2(14) - 3(1)
a = 60 - 28 - 3
a = 29
So, we have a = 29, b = 14, and c = 1.
And there you have it! The numbers that will give us the largest product are 29, 14, and 1.
Now, let's multiply them together and see what we get:
29 * 14 * 1 = 406
Voila! The largest product we can get with these numbers is 406.
Remember, math can be a clowning adventure too!
To find the numbers that will maximize the product, let's assign variables to each number.
Let:
- The first number be x
- The second number be y
- The third number be z
We have two pieces of information:
1. The sum of the three numbers is 30:
x + y + z = 30
2. The first number plus twice the second number plus three times the third number add up to 60:
x + 2y + 3z = 60
To find the maximum product, we need to maximize the values of x, y, and z.
To solve this system of equations, we can use substitution or elimination. Let's use the substitution method:
From equation 1, we can rearrange it to express x in terms of y and z:
x = 30 - y - z
Substitute this value of x in equation 2:
30 - y - z + 2y + 3z = 60
Combine like terms:
y + 2z = 30
y = 30 - 2z
Substitute this value of y in equation 1:
x + 30 - 2z + z = 30
x - z = 0
x = z
Since we want positive numbers, let's consider the case where z > 0.
If x = z and y = 30 - 2z, the product P is given by:
P = x * y * z = (z) * (30 - 2z) * z = 30z^2 - 2z^3
Now, let's find the value of z that maximizes the product P.
To find the maximum, we need to calculate the derivative of P with respect to z and set it to zero:
dP/dz = 60z - 6z^2 = 0
Simplifying, we have:
6z(10 - z) = 0
From this equation, we have two cases to consider:
1. z = 0
If z = 0, then x = 0 and y = 30. However, we want the numbers to be positive, so this case is not valid.
2. 10 - z = 0
If 10 - z = 0, then z = 10 and x = 10 and y = 10.
Let's check the validity of this solution:
- The sum of these numbers is 10 + 10 + 10 = 30, which satisfies the first condition.
- The first number plus twice the second number plus three times the third number is 10 + 2 * 10 + 3 * 10 = 10 + 20 + 30 = 60, which satisfies the second condition.
Therefore, the numbers that maximize the product are:
- The first number (x) = 10
- The second number (y) = 10
- The third number (z) = 10
To find the three positive numbers that maximize their product, we need to use optimization techniques. Let's break down the problem into steps:
Step 1: Define the problem.
We are asked to find three positive numbers whose sum is 30, and the sum of the first number plus twice the second number plus three times the third number is 60. We need to maximize the product of these three numbers.
Let's call the three numbers a, b, and c.
Step 2: Translate the problem into mathematical equations.
From the problem statement, we have two equations:
a + b + c = 30 (Equation 1)
a + 2b + 3c = 60 (Equation 2)
Step 3: Solve the equations simultaneously.
To solve this system of equations, we can use a variety of methods, such as substitution or elimination. In this case, we will solve it using the elimination method.
Multiply Equation 1 by 2 to eliminate a:
2a + 2b + 2c = 60 (Equation 3)
Now subtract Equation 2 from Equation 3 to eliminate a:
2a + 2b + 2c - (a + 2b + 3c) = 60 - 60
a - c = 0
a = c (Equation 4)
Substitute Equation 4 into Equation 1:
a + b + a = 30
2a + b = 30 (Equation 5)
Now we have two equations:
a = c (Equation 4)
2a + b = 30 (Equation 5)
Step 4: Use optimization techniques to find the maximum product.
To maximize the product, we can consider the following:
1. Since we want to maximize the product, we need to choose the maximum possible values for a and c. Since a = c from Equation 4, we can assume a = c = 15.
2. Substitute a = c = 15 into Equation 5:
2 * 15 + b = 30
30 + b = 30
b = 0
So our three numbers are a = 15, b = 0, and c = 15.
Step 5: Check the solution.
To ensure that these numbers satisfy all the conditions, let's verify:
a + b + c = 15 + 0 + 15 = 30 ✔️ (sum is 30)
a + 2b + 3c = 15 + 2(0) + 3(15) = 15 + 0 + 45 = 60 ✔️ (sum is 60)
Step 6: Calculate the maximum product.
The product of these three numbers is:
a * b * c = 15 * 0 * 15 = 0
Therefore, the maximum product of the three numbers that satisfy the given conditions is 0.
So the numbers that maximize the product are a = 15, b = 0, and c = 15.