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### Part (a): Show that the area enclosed is given by [tex]\( A = -2x^2 + 40x \, \text{m}^2 \)[/tex].
1. Understanding the Problem:
- We have a rectangular garden plot with one side against a brick wall.
- The gardener has 40 meters of fencing to enclose the other three sides of the rectangle.
2. Express the Length:
- Let [tex]\( x \)[/tex] be the width of the garden (perpendicular to the wall).
- The other two sides of the rectangle, which are both parallel to the wall, will add up to the remaining length of the fencing. Denote the length of the garden by [tex]\( y \)[/tex].
3. Fencing Constraint:
- The total fencing used is the sum of the lengths of the three sides: [tex]\( x + 2y = 40 \)[/tex].
- Therefore, we can express [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex]: [tex]\( y = \frac{40 - x}{2} \)[/tex].
4. Area Expression:
- The area [tex]\( A \)[/tex] of the rectangle is given by the product of its length and width: [tex]\( A = x \cdot y \)[/tex].
- Substituting [tex]\( y \)[/tex] from the fencing constraint, we get:
[tex]\[ A = x \cdot \left( \frac{40 - x}{2} \right) \][/tex]
- Simplifying this expression, we get:
[tex]\[ A = \frac{40x - x^2}{2} \][/tex]
[tex]\[ A = 20x - \frac{x^2}{2} \][/tex]
Multiplying through by 2 to simplify:
[tex]\[ A = -2x^2 + 40x \, \text{m}^2 \][/tex]
### Part (b): Find [tex]\( x \)[/tex] such that the vegetable garden has the maximum possible area.
1. Area Function:
- We have the area function: [tex]\( A = -2x^2 + 40x \)[/tex].
2. Finding Critical Points:
- To find the value of [tex]\( x \)[/tex] that maximizes the area, we can take the derivative of [tex]\( A \)[/tex] and set it to zero. This will give us the critical points:
[tex]\[ \frac{dA}{dx} = \frac{d}{dx}(-2x^2 + 40x) = -4x + 40 \][/tex]
Set the derivative equal to zero:
[tex]\[ -4x + 40 = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ -4x = -40 \][/tex]
[tex]\[ x = 10 \][/tex]
### Part (c): Find the maximum possible area.
1. Maximum Area:
- Substitute [tex]\( x \)[/tex] back into the area function to find the maximum area:
[tex]\[ A = -2(10)^2 + 40(10) \][/tex]
[tex]\[ A = -2(100) + 400 \][/tex]
[tex]\[ A = -200 + 400 \][/tex]
[tex]\[ A = 200 \][/tex]
Therefore:
- The value of [tex]\( x \)[/tex] that maximizes the area is [tex]\( 10 \, \text{m} \)[/tex].
- The maximum possible area is [tex]\( 200 \, \text{m}^2 \)[/tex].
### Part (a): Show that the area enclosed is given by [tex]\( A = -2x^2 + 40x \, \text{m}^2 \)[/tex].
1. Understanding the Problem:
- We have a rectangular garden plot with one side against a brick wall.
- The gardener has 40 meters of fencing to enclose the other three sides of the rectangle.
2. Express the Length:
- Let [tex]\( x \)[/tex] be the width of the garden (perpendicular to the wall).
- The other two sides of the rectangle, which are both parallel to the wall, will add up to the remaining length of the fencing. Denote the length of the garden by [tex]\( y \)[/tex].
3. Fencing Constraint:
- The total fencing used is the sum of the lengths of the three sides: [tex]\( x + 2y = 40 \)[/tex].
- Therefore, we can express [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex]: [tex]\( y = \frac{40 - x}{2} \)[/tex].
4. Area Expression:
- The area [tex]\( A \)[/tex] of the rectangle is given by the product of its length and width: [tex]\( A = x \cdot y \)[/tex].
- Substituting [tex]\( y \)[/tex] from the fencing constraint, we get:
[tex]\[ A = x \cdot \left( \frac{40 - x}{2} \right) \][/tex]
- Simplifying this expression, we get:
[tex]\[ A = \frac{40x - x^2}{2} \][/tex]
[tex]\[ A = 20x - \frac{x^2}{2} \][/tex]
Multiplying through by 2 to simplify:
[tex]\[ A = -2x^2 + 40x \, \text{m}^2 \][/tex]
### Part (b): Find [tex]\( x \)[/tex] such that the vegetable garden has the maximum possible area.
1. Area Function:
- We have the area function: [tex]\( A = -2x^2 + 40x \)[/tex].
2. Finding Critical Points:
- To find the value of [tex]\( x \)[/tex] that maximizes the area, we can take the derivative of [tex]\( A \)[/tex] and set it to zero. This will give us the critical points:
[tex]\[ \frac{dA}{dx} = \frac{d}{dx}(-2x^2 + 40x) = -4x + 40 \][/tex]
Set the derivative equal to zero:
[tex]\[ -4x + 40 = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ -4x = -40 \][/tex]
[tex]\[ x = 10 \][/tex]
### Part (c): Find the maximum possible area.
1. Maximum Area:
- Substitute [tex]\( x \)[/tex] back into the area function to find the maximum area:
[tex]\[ A = -2(10)^2 + 40(10) \][/tex]
[tex]\[ A = -2(100) + 400 \][/tex]
[tex]\[ A = -200 + 400 \][/tex]
[tex]\[ A = 200 \][/tex]
Therefore:
- The value of [tex]\( x \)[/tex] that maximizes the area is [tex]\( 10 \, \text{m} \)[/tex].
- The maximum possible area is [tex]\( 200 \, \text{m}^2 \)[/tex].
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