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Algebra

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Help I dont understand

Let x and y be nonnegative real numbers.  If x^2 + 3y^2 = 18, then find the maximum value of x + y.

Dec 19, 2023

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We can solve for y in terms of x using the given equation:

y^2 = (18 - x^2) / 3

Taking the square root of both sides (and noting that we only want nonnegative values of y):

y = sqrt((18 - x^2) / 3)

Now, we want to maximize x + y subject to the condition x^2 + 3y^2 = 18. To do this, we can use Lagrange multipliers.

Define the function:

L(x, y, λ) = x + y - λ(x^2 + 3y^2 - 18)

Taking partial derivatives with respect to x, y, and λ:

∂L/∂x = 1 - 2λx = 0 ∂L/∂y = 1 - 6λy = 0 ∂L/∂λ = -(x^2 + 3y^2 - 18) = 0

From the first equation, 1 - 2λx = 0, we get λ = 1/2x. Substituting this into the second equation, 1 - 6λy = 0, we get:

1 - 3x/y = 0

Solving for y, we get:

y = 3x / (1 - 3x)

Substituting this back into the equation x^2 + 3y^2 = 18, we get:

x^2 + 3 * 3x^2 / (1 - 3x)^2 = 18

Simplifying this equation, we get:

12x^4 - 54x^3 + 141x^2 - 108x + 144 = 0

This factors as:

(3x - 4)(4x^3 - 27x^2 + 36x - 36) = 0

We are only interested in the positive values of x, so we focus on the first factor:

3x - 4 = 0

x = 4/3

Substituting this back into the equation y = 3x / (1 - 3x), we get:

y = 4 / (1 - 3 * 4/3) = 4

Therefore, the maximum value of x + y is 4/3 + 4 = 16/3.

In conclusion, the maximum value of x + y for non-negative real numbers x and y satisfying the equation x^2 + 3y^2 = 18 is 16/3.

Dec 19, 2023