Discover new perspectives and gain insights with IDNLearn.com. Find accurate and detailed answers to your questions from our experienced and dedicated community members.
Sagot :
### 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]
Thank you for being part of this discussion. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Thank you for visiting IDNLearn.com. We’re here to provide dependable answers, so visit us again soon.
