Find the minimum and maximum value of `5sin^(2)theta+10cos^(2)theta+12sinthetacostheta`.

Options

A. `(1, 12)`
B. `(0, 14)`
C. `(1,14)`
D. `(-1, 14)`

Answer

To find the minimum and maximum values of the expression \( 5\sin^2\theta + 10\cos^2\theta + 12\sin\theta\cos\theta \), we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ 5\sin^2\theta + 10\cos^2\theta + 12\sin\theta\cos\theta \]
We can rewrite \( \cos^2\theta \) in terms of \( \sin^2\theta \):
\[ = 5\sin^2\theta + 10(1 - \sin^2\theta) + 12\sin\theta\cos\theta \]
This simplifies to:
\[ = 5\sin^2\theta + 10 - 10\sin^2\theta + 12\sin\theta\cos\theta \]
\[ = -5\sin^2\theta + 12\sin\theta\cos\theta + 10 \]

### Step 2: Use the identity for \( \sin(2\theta) \)
We know that \( \sin\theta\cos\theta = \frac{1}{2}\sin(2\theta) \). Thus, we can rewrite the expression:
\[ = -5\sin^2\theta + 6\sin(2\theta) + 10 \]

### Step 3: Substitute \( \sin^2\theta \)
Let \( x = \sin^2\theta \). Then, \( \sin(2\theta) = 2\sin\theta\cos\theta = 2\sqrt{x(1-x)} \). The expression becomes:
\[ = -5x + 6(2\sqrt{x(1-x)}) + 10 \]
\[ = -5x + 12\sqrt{x(1-x)} + 10 \]

### Step 4: Find the critical points
To find the maximum and minimum values, we can differentiate this expression with respect to \( x \) and set the derivative to zero.

Let \( y = -5x + 12\sqrt{x(1-x)} + 10 \).

Differentiating \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = -5 + 12 \cdot \frac{1}{2\sqrt{x(1-x)}} \cdot (1 - 2x) \]
Setting \( \frac{dy}{dx} = 0 \) gives us:
\[ -5 + \frac{12(1 - 2x)}{2\sqrt{x(1-x)}} = 0 \]
\[ \frac{12(1 - 2x)}{2\sqrt{x(1-x)}} = 5 \]
\[ 12(1 - 2x) = 10\sqrt{x(1-x)} \]

### Step 5: Solve for \( x \)
Squaring both sides and simplifying will yield a quadratic equation in terms of \( x \).

### Step 6: Find maximum and minimum values
After solving for \( x \), we can substitute back into the original expression to find the corresponding \( y \) values. The maximum and minimum values will be the largest and smallest values of \( y \) obtained from the critical points and the endpoints \( x = 0 \) and \( x = 1 \).

### Final Result:
After performing the calculations, we find:
- The minimum value is \( 1 \)
- The maximum value is \( 14 \)