Let $x$ and $y$ be nonnegative real numbers. If $x^2 + 5y^2 = 30$, then find the maximum value of $x + y$.

1 answer

To maximize \( z = x + y \) subject to the constraint \( x^2 + 5y^2 = 30 \), we can use the method of Lagrange multipliers or apply a substitution based on the constraint.

First, we solve the constraint for \( y^2 \) in terms of \( x \):

\[ 5y^2 = 30 - x^2 \quad \Rightarrow \quad y^2 = \frac{30 - x^2}{5} \]

Next, consider substituting \( y \) back into \( z \):

\[ y = \sqrt{\frac{30 - x^2}{5}} \]

We can now express \( z \) solely in terms of \( x \):

\[ z = x + \sqrt{\frac{30 - x^2}{5}} \]

Now, we will differentiate \( z \) with respect to \( x \) and find the critical points. First, calculate \( z^2 \) for easier differentiation:

\[ z = x + \sqrt{\frac{30 - x^2}{5}} \]

Rewriting gives us:

\[ z = x + \frac{1}{\sqrt{5}} \sqrt{30 - x^2} \]

Let's square \( z \):

\[ z^2 = \left( x + \frac{1}{\sqrt{5}} \sqrt{30 - x^2} \right)^2 \]

For maximization, let's differentiate \( z \):

\[ \frac{dz}{dx} = 1 + \frac{1}{\sqrt{5}} \cdot \frac{1}{2\sqrt{30 - x^2}} \cdot (-2x) \]

Setting the derivative equal to zero for critical points:

\[ 1 - \frac{x}{\sqrt{5(30 - x^2)}} = 0 \]

This implies:

\[ \frac{x}{\sqrt{5(30 - x^2)}} = 1 \]

Squaring both sides gives:

\[ x^2 = 5(30 - x^2) \]

This simplifies to:

\[ x^2 + 5x^2 = 150 \quad \Rightarrow \quad 6x^2 = 150 \quad \Rightarrow \quad x^2 = 25 \quad \Rightarrow \quad x = 5 \]

Substituting \( x = 5 \) back into the constraint to find \( y \):

\[ 5^2 + 5y^2 = 30 \quad \Rightarrow \quad 25 + 5y^2 = 30 \quad \Rightarrow \quad 5y^2 = 5 \quad \Rightarrow \quad y^2 = 1 \quad \Rightarrow \quad y = 1 \]

Finally, the maximum value of \( x + y \) is:

\[ x + y = 5 + 1 = 6 \]

To ensure this is indeed the maximum, we check the endpoints. If \( x = 0 \):

\[ 5y^2 = 30 \quad \Rightarrow \quad y^2 = 6 \quad \Rightarrow \quad y = \sqrt{6} \quad \Rightarrow \quad z = \sqrt{6} \approx 2.45 \]

If \( y = 0 \):

\[ x^2 = 30 \quad \Rightarrow \quad x = \sqrt{30} \quad \Rightarrow \quad z = \sqrt{30} \approx 5.48 \]

Thus, the maximum occurs at \( (5, 1) \).

Thus, the maximum value of \( x + y \) is:

\[ \boxed{6} \]