Suppose that the temperature T at every point (x,y) in the plane cartesian is given by the formula `T=1-x^(2)+2y^(2)`. The correct statement about the maximum and minimum temperature along the line x+y=1 is

Options

A. Minimum is -1. There is no maximum
B. Maximum is -1. There is no minimum
C. Maximum is 0. Minimum is -1
D. There is neither a maximum nor a minimum along the line

Answer

To find the maximum and minimum temperature along the line \( x + y = 1 \) given the temperature function \( T = 1 - x^2 + 2y^2 \), we can follow these steps:

### Step 1: Substitute \( y \) in terms of \( x \)
Given the line equation \( x + y = 1 \), we can express \( y \) as:
\[ y = 1 - x \]

### Step 2: Substitute \( y \) into the temperature function
Now, we substitute \( y = 1 - x \) into the temperature function:
\[ T = 1 - x^2 + 2(1 - x)^2 \]

### Step 3: Expand the equation
Now, we expand the equation:
\[ T = 1 - x^2 + 2(1 - 2x + x^2) \]
\[ T = 1 - x^2 + 2 - 4x + 2x^2 \]
\[ T = 3 + x^2 - 4x \]

### Step 4: Rearrange the equation
Rearranging the equation gives:
\[ T = x^2 - 4x + 3 \]

### Step 5: Complete the square
To find the minimum and maximum values, we can complete the square:
\[ T = (x - 2)^2 - 1 \]

### Step 6: Analyze the completed square
From the completed square form \( T = (x - 2)^2 - 1 \):
- The term \( (x - 2)^2 \) is always non-negative and reaches its minimum value of 0 when \( x = 2 \).
- Therefore, the minimum value of \( T \) occurs at \( x = 2 \):
\[ T_{\text{min}} = 0 - 1 = -1 \]
- Since \( (x - 2)^2 \) can increase indefinitely, there is no maximum temperature along the line.

### Conclusion
Thus, the minimum temperature along the line \( x + y = 1 \) is \( -1 \), and there is no maximum temperature.

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