Find two numbers x, and y that add to 30, and for which 2x^2 +5y^2 is a minimum.

x+y = 30 or y = 30-x

let S = 2x^2 + 5y^2
= 2x^2 + 5(30-x)^2
= 2x^2 + 4500 - 300x + 5x^2
= 7x^2 - 300x + 4500
dS/dx = 14x - 300
= 0 for a max/min
x = 300/14 , sub into y = 30-x

Why did the number go to therapy?

Because it had too many negative signs and needed some positive solutions!

Alright, let's solve this math problem with a dash of humor. So, we're looking for two numbers, x and y, that add up to 30 and minimize the expression 2x^2 + 5y^2.

To make things easier, let's start by expressing one number in terms of the other. Let's say x = 30 - y.

Substituting this value into the expression, we get 2(30 - y)^2 + 5y^2.

Now, let's simplify this expression.

Expanding (30 - y)^2, we get (900 - 60y + y^2).

So, our expression becomes 2(900 - 60y + y^2) + 5y^2.

Simplifying further, we have 1800 - 120y + 2y^2 + 5y^2.

Combining like terms, we get 7y^2 - 120y + 1800.

To find the minimum value, we can use the vertex formula, which is x = -b/2a. In this case, we have a = 7 and b = -120.

Substituting these values, we get x = -(-120)/(2*7) = 120/14 = 60/7.

So, one of the numbers is x = 60/7, and the other number is y = 30 - x = 30 - (60/7).

Now, let's put the "fun" in "function" and calculate the exact values.

x = 60/7 ≈ 8.57
y = 30 - (60/7) ≈ 4.29

Therefore, the two numbers that add to 30 and minimize the expression 2x^2 + 5y^2 are approximately x = 8.57 and y = 4.29. Keep in mind that these values are rounded for simplicity. Enjoy your mathematical clowning around!

To find two numbers, x and y, that add to 30 and minimize the expression 2x^2 + 5y^2, we can use the method of differentiation.

Let's proceed step-by-step:

Step 1: Express y in terms of x using the given constraint that x + y = 30.
y = 30 - x

Step 2: Substitute the expression for y in terms of x into the objective function 2x^2 + 5y^2.
f(x) = 2x^2 + 5(30 - x)^2

Step 3: Expand the expression and simplify.
f(x) = 2x^2 + 5(900 - 60x + x^2)
= 2x^2 + 4500 - 300x + 5x^2
= 7x^2 - 300x + 4500

Step 4: To find the minimum value of f(x), we can differentiate it with respect to x and set it equal to zero.
f'(x) = 14x - 300
14x - 300 = 0

Step 5: Solve for x.
14x = 300
x = 300 / 14
x ≈ 21.43

Step 6: Substitute the value of x back into the equation x + y = 30 to find y.
21.43 + y = 30
y ≈ 8.57

Therefore, the two numbers that add to 30 and minimize the expression 2x^2 + 5y^2 are approximately x ≈ 21.43 and y ≈ 8.57.

To find the two numbers x and y that add up to 30 and minimize the expression 2x^2 + 5y^2, we can use calculus. Let's break down the problem into smaller steps:

Step 1: Express one of the variables in terms of the other variable.
Since we know that x + y = 30, we can express y as y = 30 - x.

Step 2: Rewrite the expression 2x^2 + 5y^2 in terms of a single variable.
Substituting the equation from Step 1 into the expression, we get:
2x^2 + 5(30 - x)^2.

Step 3: Expand and simplify the expression.
Expanding the squared term, we have:
2x^2 + 5(900 - 60x + x^2).
Simplifying further, we get:
2x^2 + 4500 - 300x + 5x^2.

Step 4: Find the derivative of the expression with respect to x.
To minimize the expression, we need to find where its derivative is equal to zero.
Taking the derivative of the expression, we get:
d/dx (2x^2 + 4500 - 300x + 5x^2) = 4x - 300 + 10x.

Step 5: Set the derivative equal to zero and solve for x.
Setting 4x - 300 + 10x = 0 and solving for x gives us:
14x - 300 = 0,
14x = 300,
x = 300/14 = 21.43 (rounded to two decimal places).

Step 6: Substitute the value of x back into the equation to find y.
Using y = 30 - x, we substitute x = 21.43:
y = 30 - 21.43 = 8.57 (rounded to two decimal places).

Therefore, the two numbers that add up to 30 and minimize the expression 2x^2 + 5y^2 are x = 21.43 and y = 8.57.