Answer :
To find the critical points, domain endpoints, and local extreme values for the function
[tex]\[ y = 6 x \sqrt{100 - x^2} \][/tex]
follow these steps:
### 1. Define the Function and its Domain
We start with:
[tex]\[ y = 6 x \sqrt{100 - x^2} \][/tex]
The domain of [tex]\( \sqrt{100 - x^2} \)[/tex] requires:
[tex]\[ 100 - x^2 \geq 0 \implies -10 \leq x \leq 10 \][/tex]
### 2. Find Critical Points
Critical points occur where the first derivative is zero or undefined. Let's find the first derivative [tex]\( y' \)[/tex]:
[tex]\[ y' = \frac{d}{dx} \left( 6 x \sqrt{100 - x^2} \right) \][/tex]
We can use the product rule:
[tex]\[ y' = 6 \left( \frac{d}{dx} x \cdot \sqrt{100 - x^2} + x \cdot \frac{d}{dx} \sqrt{100 - x^2} \right) \][/tex]
First, calculate each part:
- [tex]\(\frac{d}{dx} x = 1\)[/tex]
- [tex]\(\frac{d}{dx} \sqrt{100 - x^2} = \frac{-x}{\sqrt{100 - x^2}} \)[/tex]
Now, combine them:
[tex]\[ y' = 6 \left( \sqrt{100 - x^2} + x \cdot \frac{-x}{\sqrt{100 - x^2}} \right) \][/tex]
[tex]\[ y' = 6 \left( \sqrt{100 - x^2} - \frac{x^2}{\sqrt{100 - x^2}} \right) \][/tex]
[tex]\[ y' = 6 \left( \frac{100 - x^2 - x^2}{\sqrt{100 - x^2}} \right) \][/tex]
[tex]\[ y' = 6 \left( \frac{100 - 2x^2}{\sqrt{100 - x^2}} \right) \][/tex]
Set the first derivative to 0 to find critical points:
[tex]\[ 6 \left( \frac{100 - 2x^2}{\sqrt{100 - x^2}} \right) = 0 \][/tex]
[tex]\[ 100 - 2x^2 = 0 \][/tex]
[tex]\[ 2x^2 = 100 \][/tex]
[tex]\[ x^2 = 50 \][/tex]
[tex]\[ x = \pm \sqrt{50} = \pm 5\sqrt{2} \][/tex]
### 3. Check the Domain Endpoints
The domain endpoints are:
[tex]\[ x = 10 \quad \text{and} \quad x = -10 \][/tex]
### 4. Evaluate the Function at These Points
Evaluate [tex]\( y \)[/tex] at the critical points and domain endpoints:
[tex]\[ y(5\sqrt{2}) = 6 \cdot 5\sqrt{2} \cdot \sqrt{100 - (5\sqrt{2})^2} \][/tex]
[tex]\[ = 6 \cdot 5\sqrt{2} \cdot \sqrt{100 - 50} \][/tex]
[tex]\[ = 6 \cdot 5\sqrt{2} \cdot \sqrt{50} \][/tex]
[tex]\[ = 6 \cdot 5\sqrt{2} \cdot 5\sqrt{2} \][/tex]
[tex]\[ = 6 \cdot 25 \cdot 2 = 300 \][/tex]
[tex]\[ y(-5\sqrt{2}) = 6 \cdot (-5\sqrt{2}) \cdot \sqrt{100 - 50} \][/tex]
[tex]\[ = 6 \cdot (-5\sqrt{2}) \cdot \sqrt{50} \][/tex]
[tex]\[ = 6 \cdot (-5\sqrt{2}) \cdot 5\sqrt{2} \][/tex]
[tex]\[ = 6 \cdot (-25) \cdot 2 = -300 \][/tex]
Domain endpoints:
[tex]\[ y(10) = 6 \cdot 10 \cdot \sqrt{100 - 100} = 6 \cdot 10 \cdot 0 = 0 \][/tex]
[tex]\[ y(-10) = 6 \cdot (-10) \cdot \sqrt{100 - 100} = 6 \cdot (-10) \cdot 0 = 0 \][/tex]
### 5. Determine the Local Maxima
- At [tex]\( x = 5\sqrt{2} \)[/tex], [tex]\( y = 300 \)[/tex]
- At [tex]\( x = -5\sqrt{2} \)[/tex], [tex]\( y = -300 \)[/tex]
- At [tex]\( x = 10 \)[/tex] and [tex]\( x = -10 \)[/tex], [tex]\( y = 0 \)[/tex]
### Conclusion
The local maximum is at [tex]\( x = 5\sqrt{2} \)[/tex] because it gives the highest value:
[tex]\[ \boxed{(5\sqrt{2}, 300)} \][/tex]
This is the only local maximum among the points evaluated. Therefore, the correct choice is:
A. The point(s) corresponding to the local maxima is/are [tex]\((5\sqrt{2}, 300)\)[/tex]
[tex]\[ y = 6 x \sqrt{100 - x^2} \][/tex]
follow these steps:
### 1. Define the Function and its Domain
We start with:
[tex]\[ y = 6 x \sqrt{100 - x^2} \][/tex]
The domain of [tex]\( \sqrt{100 - x^2} \)[/tex] requires:
[tex]\[ 100 - x^2 \geq 0 \implies -10 \leq x \leq 10 \][/tex]
### 2. Find Critical Points
Critical points occur where the first derivative is zero or undefined. Let's find the first derivative [tex]\( y' \)[/tex]:
[tex]\[ y' = \frac{d}{dx} \left( 6 x \sqrt{100 - x^2} \right) \][/tex]
We can use the product rule:
[tex]\[ y' = 6 \left( \frac{d}{dx} x \cdot \sqrt{100 - x^2} + x \cdot \frac{d}{dx} \sqrt{100 - x^2} \right) \][/tex]
First, calculate each part:
- [tex]\(\frac{d}{dx} x = 1\)[/tex]
- [tex]\(\frac{d}{dx} \sqrt{100 - x^2} = \frac{-x}{\sqrt{100 - x^2}} \)[/tex]
Now, combine them:
[tex]\[ y' = 6 \left( \sqrt{100 - x^2} + x \cdot \frac{-x}{\sqrt{100 - x^2}} \right) \][/tex]
[tex]\[ y' = 6 \left( \sqrt{100 - x^2} - \frac{x^2}{\sqrt{100 - x^2}} \right) \][/tex]
[tex]\[ y' = 6 \left( \frac{100 - x^2 - x^2}{\sqrt{100 - x^2}} \right) \][/tex]
[tex]\[ y' = 6 \left( \frac{100 - 2x^2}{\sqrt{100 - x^2}} \right) \][/tex]
Set the first derivative to 0 to find critical points:
[tex]\[ 6 \left( \frac{100 - 2x^2}{\sqrt{100 - x^2}} \right) = 0 \][/tex]
[tex]\[ 100 - 2x^2 = 0 \][/tex]
[tex]\[ 2x^2 = 100 \][/tex]
[tex]\[ x^2 = 50 \][/tex]
[tex]\[ x = \pm \sqrt{50} = \pm 5\sqrt{2} \][/tex]
### 3. Check the Domain Endpoints
The domain endpoints are:
[tex]\[ x = 10 \quad \text{and} \quad x = -10 \][/tex]
### 4. Evaluate the Function at These Points
Evaluate [tex]\( y \)[/tex] at the critical points and domain endpoints:
[tex]\[ y(5\sqrt{2}) = 6 \cdot 5\sqrt{2} \cdot \sqrt{100 - (5\sqrt{2})^2} \][/tex]
[tex]\[ = 6 \cdot 5\sqrt{2} \cdot \sqrt{100 - 50} \][/tex]
[tex]\[ = 6 \cdot 5\sqrt{2} \cdot \sqrt{50} \][/tex]
[tex]\[ = 6 \cdot 5\sqrt{2} \cdot 5\sqrt{2} \][/tex]
[tex]\[ = 6 \cdot 25 \cdot 2 = 300 \][/tex]
[tex]\[ y(-5\sqrt{2}) = 6 \cdot (-5\sqrt{2}) \cdot \sqrt{100 - 50} \][/tex]
[tex]\[ = 6 \cdot (-5\sqrt{2}) \cdot \sqrt{50} \][/tex]
[tex]\[ = 6 \cdot (-5\sqrt{2}) \cdot 5\sqrt{2} \][/tex]
[tex]\[ = 6 \cdot (-25) \cdot 2 = -300 \][/tex]
Domain endpoints:
[tex]\[ y(10) = 6 \cdot 10 \cdot \sqrt{100 - 100} = 6 \cdot 10 \cdot 0 = 0 \][/tex]
[tex]\[ y(-10) = 6 \cdot (-10) \cdot \sqrt{100 - 100} = 6 \cdot (-10) \cdot 0 = 0 \][/tex]
### 5. Determine the Local Maxima
- At [tex]\( x = 5\sqrt{2} \)[/tex], [tex]\( y = 300 \)[/tex]
- At [tex]\( x = -5\sqrt{2} \)[/tex], [tex]\( y = -300 \)[/tex]
- At [tex]\( x = 10 \)[/tex] and [tex]\( x = -10 \)[/tex], [tex]\( y = 0 \)[/tex]
### Conclusion
The local maximum is at [tex]\( x = 5\sqrt{2} \)[/tex] because it gives the highest value:
[tex]\[ \boxed{(5\sqrt{2}, 300)} \][/tex]
This is the only local maximum among the points evaluated. Therefore, the correct choice is:
A. The point(s) corresponding to the local maxima is/are [tex]\((5\sqrt{2}, 300)\)[/tex]
