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### Solution:
#### Part 10.1: Expressing the Area of Rectangle PQRS
To express the area of the rectangle PQRS:
1. Let the coordinates of point [tex]\( P \)[/tex] be [tex]\( (x, h(x)) \)[/tex], where [tex]\( h(x) \)[/tex] is given by the function [tex]\( h(x) = x^2 \)[/tex].
2. Given that the x-axis and the line [tex]\( x = 6 \)[/tex] are boundaries, the other corner [tex]\( Q \)[/tex] of the rectangle on the x-axis will be [tex]\( (x, 0) \)[/tex], and the width of the rectangle is from [tex]\( x = 6 \)[/tex] to [tex]\( x \)[/tex].
Thus, the width of the rectangle [tex]\( w \)[/tex] is:
[tex]\[ w = 6 - x \][/tex]
The height of the rectangle [tex]\( h \)[/tex] at point [tex]\( P \)[/tex] is:
[tex]\[ h = h(x) = x^2 \][/tex]
Thus, the area [tex]\( A \)[/tex] of the rectangle PQRS is given by:
[tex]\[ A = \text{width} \times \text{height} \][/tex]
[tex]\[ A = (6 - x) \times x^2 \][/tex]
[tex]\[ A = 6x^2 - x^3 \][/tex]
Therefore, we have expressed the area [tex]\( A \)[/tex] as:
[tex]\[ A = 6x^2 - x^3 \][/tex]
#### Part 10.2: Determining the Largest Possible Area
Next, to find the largest possible area for the rectangle PQRS, we need to maximize the area function [tex]\( A = 6x^2 - x^3 \)[/tex].
1. First Derivative:
To find the critical points, we first take the derivative of [tex]\( A \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{dA}{dx} = \frac{d}{dx}(6x^2 - x^3) \][/tex]
[tex]\[ \frac{dA}{dx} = 12x - 3x^2 \][/tex]
2. Setting the derivative to zero:
We set the first derivative to zero to find the critical points:
[tex]\[ 12x - 3x^2 = 0 \][/tex]
[tex]\[ 3x(4 - x) = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ x = 0 \quad \text{or} \quad x = 4 \][/tex]
3. Second Derivative:
To confirm which critical point gives the maximum area, we take the second derivative of [tex]\( A \)[/tex]:
[tex]\[ \frac{d^2A}{dx^2} = \frac{d}{dx}(12x - 3x^2) \][/tex]
[tex]\[ \frac{d^2A}{dx^2} = 12 - 6x \][/tex]
4. Evaluating the second derivative at the critical points:
- At [tex]\( x = 0 \)[/tex]:
[tex]\[ \frac{d^2A}{dx^2} \Bigg|_{x=0} = 12 - 6(0) = 12 \][/tex]
Since [tex]\( \frac{d^2A}{dx^2} > 0 \)[/tex], [tex]\( x = 0 \)[/tex] is a local minimum.
- At [tex]\( x = 4 \)[/tex]:
[tex]\[ \frac{d^2A}{dx^2} \Bigg|_{x=4} = 12 - 6(4) = 12 - 24 = -12 \][/tex]
Since [tex]\( \frac{d^2A}{dx^2} < 0 \)[/tex], [tex]\( x = 4 \)[/tex] is a local maximum.
Therefore, the maximum area is achieved at [tex]\( x = 4 \)[/tex].
5. Calculating the maximum area:
Substituting [tex]\( x = 4 \)[/tex] into the area function:
[tex]\[ A = 6x^2 - x^3 \][/tex]
[tex]\[ A = 6(4)^2 - (4)^3 \][/tex]
[tex]\[ A = 6(16) - 64 \][/tex]
[tex]\[ A = 96 - 64 \][/tex]
[tex]\[ A = 32 \][/tex]
The largest possible area for the rectangle PQRS is:
[tex]\[ \boxed{32} \][/tex]
#### Part 10.1: Expressing the Area of Rectangle PQRS
To express the area of the rectangle PQRS:
1. Let the coordinates of point [tex]\( P \)[/tex] be [tex]\( (x, h(x)) \)[/tex], where [tex]\( h(x) \)[/tex] is given by the function [tex]\( h(x) = x^2 \)[/tex].
2. Given that the x-axis and the line [tex]\( x = 6 \)[/tex] are boundaries, the other corner [tex]\( Q \)[/tex] of the rectangle on the x-axis will be [tex]\( (x, 0) \)[/tex], and the width of the rectangle is from [tex]\( x = 6 \)[/tex] to [tex]\( x \)[/tex].
Thus, the width of the rectangle [tex]\( w \)[/tex] is:
[tex]\[ w = 6 - x \][/tex]
The height of the rectangle [tex]\( h \)[/tex] at point [tex]\( P \)[/tex] is:
[tex]\[ h = h(x) = x^2 \][/tex]
Thus, the area [tex]\( A \)[/tex] of the rectangle PQRS is given by:
[tex]\[ A = \text{width} \times \text{height} \][/tex]
[tex]\[ A = (6 - x) \times x^2 \][/tex]
[tex]\[ A = 6x^2 - x^3 \][/tex]
Therefore, we have expressed the area [tex]\( A \)[/tex] as:
[tex]\[ A = 6x^2 - x^3 \][/tex]
#### Part 10.2: Determining the Largest Possible Area
Next, to find the largest possible area for the rectangle PQRS, we need to maximize the area function [tex]\( A = 6x^2 - x^3 \)[/tex].
1. First Derivative:
To find the critical points, we first take the derivative of [tex]\( A \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{dA}{dx} = \frac{d}{dx}(6x^2 - x^3) \][/tex]
[tex]\[ \frac{dA}{dx} = 12x - 3x^2 \][/tex]
2. Setting the derivative to zero:
We set the first derivative to zero to find the critical points:
[tex]\[ 12x - 3x^2 = 0 \][/tex]
[tex]\[ 3x(4 - x) = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ x = 0 \quad \text{or} \quad x = 4 \][/tex]
3. Second Derivative:
To confirm which critical point gives the maximum area, we take the second derivative of [tex]\( A \)[/tex]:
[tex]\[ \frac{d^2A}{dx^2} = \frac{d}{dx}(12x - 3x^2) \][/tex]
[tex]\[ \frac{d^2A}{dx^2} = 12 - 6x \][/tex]
4. Evaluating the second derivative at the critical points:
- At [tex]\( x = 0 \)[/tex]:
[tex]\[ \frac{d^2A}{dx^2} \Bigg|_{x=0} = 12 - 6(0) = 12 \][/tex]
Since [tex]\( \frac{d^2A}{dx^2} > 0 \)[/tex], [tex]\( x = 0 \)[/tex] is a local minimum.
- At [tex]\( x = 4 \)[/tex]:
[tex]\[ \frac{d^2A}{dx^2} \Bigg|_{x=4} = 12 - 6(4) = 12 - 24 = -12 \][/tex]
Since [tex]\( \frac{d^2A}{dx^2} < 0 \)[/tex], [tex]\( x = 4 \)[/tex] is a local maximum.
Therefore, the maximum area is achieved at [tex]\( x = 4 \)[/tex].
5. Calculating the maximum area:
Substituting [tex]\( x = 4 \)[/tex] into the area function:
[tex]\[ A = 6x^2 - x^3 \][/tex]
[tex]\[ A = 6(4)^2 - (4)^3 \][/tex]
[tex]\[ A = 6(16) - 64 \][/tex]
[tex]\[ A = 96 - 64 \][/tex]
[tex]\[ A = 32 \][/tex]
The largest possible area for the rectangle PQRS is:
[tex]\[ \boxed{32} \][/tex]
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