Find the maximum and minimum values of the function $$ f(x, y)=2 x^{2}+3 y^{2}-4 x-5 $$ on the domain $$ x^{2}+y^{2} \leq 25 $$. The maximum value of $$ f(x, y) $$ is: List the point(s) where the function attains its maximum as an ordered pair, such as $$ (-6,3) $$, or a list of ordered pairs if there is more than one point, such as $$ (1,3),(-4,7) $$. The minimum value of $$ f(x, y) $$ is: -7 List points where the function attains its minimum as an ordered pair, such as $$ (-6,3) $$, or a list of ordered pairs if there is more than one point, such as $$ (1,3),(-4,7) $$. $$ (1,0) $$

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To find the maximum and minimum values of the function

$$
f(x, y) = 2x^2 + 3y^2 - 4x - 5
$$

on the domain

$$
x^2 + y^2 \leq 25,
$$

we will follow these steps:

1. **Find the critical points of $$ f(x, y) $$** by taking the partial derivatives and setting them to zero.
2. **Evaluate $$ f(x, y) $$** at the critical points and on the boundary of the domain defined by $$ x^2 + y^2 = 25 $$.
3. **Compare the values** to determine the maximum and minimum.

### Step 1: Find Critical Points

First, we calculate the partial derivatives of $$ f $$:

$$
f_x = \frac{\partial f}{\partial x} = 4x - 4,
$$
$$
f_y = \frac{\partial f}{\partial y} = 6y.
$$

Setting these equal to zero gives us the equations:

$$
4x - 4 = 0 \quad \Rightarrow \quad x = 1,
$$
$$
6y = 0 \quad \Rightarrow \quad y = 0.
$$

Thus, the critical point is $$ (1, 0) $$.

### Step 2: Evaluate on the Boundary

Next, we need to evaluate $$ f(x, y) $$ on the boundary defined by $$ x^2 + y^2 = 25 $$. We can use a parameterization for the boundary:

Let $$ x = 5\cos(\theta) $$ and $$ y = 5\sin(\theta) $$ for $$ \theta \in [0, 2\pi] $$.

Substituting into $$ f $$:

$$
f(5\cos(\theta), 5\sin(\theta)) = 2(5\cos(\theta))^2 + 3(5\sin(\theta))^2 - 4(5\cos(\theta)) - 5.
$$

This simplifies to:

$$
f(5\cos(\theta), 5\sin(\theta)) = 50\cos^2(\theta) + 75\sin^2(\theta) - 20\cos(\theta) - 5.
$$

Combining terms gives:

$$
f(5\cos(\theta), 5\sin(\theta)) = 50\cos^2(\theta) + 75\sin^2(\theta) - 20\cos(\theta) - 5.
$$

### Step 3: Find Maximum and Minimum Values

To find the maximum and minimum values, we can evaluate $$ f $$ at the critical point and at several points on the boundary.

1. **At the critical point $$ (1, 0) $$**:

$$
f(1, 0) = 2(1)^2 + 3(0)^2 - 4(1) - 5 = 2 - 4 - 5 = -7.
$$

2. **On the boundary $$ x^2 + y^2 = 25 $$**, we can evaluate at key angles:

- $$ \theta = 0 $$ (point $$ (5, 0) $$):
$$
f(5, 0) = 2(5)^2 + 3(0)^2 - 4(5) - 5 = 50 - 20 - 5 = 25.
$$

- $$ \theta = \frac{\pi}{2} $$ (point $$ (0, 5) $$):
$$
f(0, 5) = 2(0)^2 + 3(5)^2 - 4(0) - 5 = 75 - 5 = 70.
$$

- $$ \theta = \pi $$ (point $$ (-5, 0) $$):
$$
f(-5, 0) = 2(-5)^2 + 3(0)^2 - 4(-5) - 5 = 50 + 20 - 5 = 65.
$$

- $$ \theta = \frac{3\pi}{2} $$ (point $$ (0, -5) $$):
$$
f(0, -5) = 2(0)^2 + 3(-5)^2 - 4(0) - 5 = 75 - 5 = 70.
$$

### Summary of Values

- At $$ (1, 0) $$: $$ f(1, 0) = -7 $$ (minimum)
- At $$ (5, 0) $$: $$ f(5, 0) = 25 $$
- At $$ (0, 5) $$: $$ f(0, 5) = 70 $$
- At $$ (-5, 0) $$: $$ f(-5, 0) = 65 $$
- At $$ (0, -5) $$: $$ f(0, -5) = 70 $$

### Conclusion

- The **maximum value** of $$ f(x, y) $$ is $$ 70 $$ at the points $$ (0, 5) $$ and $$ (0, -5) $$.
- The **minimum value** of $$ f(x, y) $$ is $$ -7 $$ at the point $$ (1, 0) $$.

Thus, the final answers are:

- Maximum value: $$ 70 $$ at points $$ (0, 5), (0, -5) $$.
- Minimum value: $$ -7 $$ at point $$ (1, 0) $$.
The maximum value of $$ f(x, y) $$ is 70 at points $$ (0, 5) $$ and $$ (0, -5) $$. The minimum value is -7 at point $$ (1, 0) $$.

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