the sumof three positive numbers is 26.the second number is 3times as large as the first.if te sum of the squares of these numbers is least find the numbers
Let's break down the information given:
- The sum of three positive numbers is 26.
- The second number is 3 times as large as the first.
To solve this problem and find the numbers, we can set up an equation. Let's assign variables to the numbers:
Let the first number be x.
The second number is then 3x, as it is 3 times as large as the first number.
Let the third number be y.
Using the given information, we can now form the equation:
x + 3x + y = 26
Simplifying the equation:
4x + y = 26
To find the sum of squares of these numbers, we need to minimize it. The sum of squares is given by:
x^2 + (3x)^2 + y^2
Simplifying the equation:
x^2 + 9x^2 + y^2
10x^2 + y^2
To find the least value for this equation, we need to minimize it. Let's solve for x and y to find the values that minimize the sum of squares.
Now, we need to find the values of x and y that satisfy the equation 4x + y = 26. There are multiple possibilities, so we will use a method called substitution to find the values that minimize the sum of squares.
Step 1: Solve the equation 4x + y = 26 for y:
y = 26 - 4x
Step 2: Substitute the value of y in terms of x into the equation 10x^2 + y^2:
10x^2 + (26 - 4x)^2
Step 3: Simplify the equation:
10x^2 + (676 - 208x + 16x^2)
Step 4: Combine like terms:
10x^2 + 16x^2 - 208x + 676
Step 5: Combine like terms:
26x^2 - 208x + 676
To find the least value for this quadratic equation, we can use the formula for the vertex of a parabola:
x = -b / (2a)
In this case, a = 26, b = -208. Plug these values into the formula:
x = -(-208) / (2 * 26) = 208 / 52 = 4
Now, substitute the value of x back into the equation 4x + y = 26 to find y:
4(4) + y = 26
16 + y = 26
y = 26 - 16
y = 10
Therefore, the numbers that minimize the sum of squares are:
- The first number (x) = 4
- The second number (3x) = 3 * 4 = 12
- The third number (y) = 10
So, the three numbers are 4, 12, and 10.
To find the numbers, let's assign variables for the three positive numbers. Let's call the first number "x," the second number "3x" (since it is 3 times as large as the first), and the third number "26 - (x + 3x)" (since the sum of all three numbers is 26).
Now let's calculate the sum of the squares of these numbers:
x^2 + (3x)^2 + (26 - (x + 3x))^2
Simplifying this expression:
x^2 + 9x^2 + (26 - 4x)^2
Expanding the squared term:
x^2 + 9x^2 + (676 - 208x + 16x^2)
Combining like terms:
10x^2 - 192x + 676
Since we are looking for the least sum of squares, we need to find the minimum point of this quadratic equation. To find the minimum point, we can use the formula x = -b/2a, where a = 10, and b = -192.
x = -(-192) / 2(10)
x = 192 / 20
x = 9.6
Since we are looking for positive whole numbers, we can round x to the nearest whole number, giving us x = 10.
Now, we can find the second number:
3x = 3(10) = 30
Finally, we can find the third number:
26 - (x + 3x) = 26 - (10 + 30) = 26 - 40 = 6
Therefore, the three numbers are 10, 30, and 6.
that sucks
x+y+z = 26
y = 3x
You want minimal x^2+y^2+z^2, so check out a few possibilities:
x y z ∑*^2
1 3 22 488
2 6 18 332
3 9 14 208
...
I think you can figure it out.