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You have 3L feet of fence to make a rectangular vegetable garden alongside the wall of your house, where L is a positive constant. The wall of the house bounds one side of the vegetable garden. What is the largest possible area for the vegetable garden?

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ajeigbeibraheem

You have 3L feet of fence to make a rectangular vegetable garden alongside the wall of your house, where L is a positive constant. The wall of the house bounds one side of the vegetable garden. The largest possible area for the vegetable garden is: [tex]\mathbf{ \dfrac{ 9L^2}{8}}[/tex]

From the information given:

  • the fencing of the length L = 3L
  • Let us recall that only one side of the vegetable garden was identified to be the wall

As such, there won't be any fencing on the other end.

  • suppose, we denote the length of the garden as = a,
  • and, the width of the garden as = w

Then;

  • 2a + w = 3L
  • w = 3L - 2a

From the formula for determining the area of a rectangle;

  • Area of rectangle = length × width
  • Area(A) = a × w
  • A = a × (3L - 2a)
  • A = 3La - 2a²

Taking the derivative of the above equation, we have:

0 = 3L - 4a

[tex]\mathbf{a = \dfrac{3L}{4}}[/tex]

Replacing the value of (a) into the formula for calculating the area of a rectangle, we have:

[tex]\mathbf{A = 3L(\dfrac{3L}{4}) - 2(\dfrac{3L}{4})^2}[/tex]

[tex]\mathbf{A = 3L(\dfrac{3L}{4}) - 2(\dfrac{9L^2}{16})}[/tex]

[tex]\mathbf{A = (\dfrac{9L^2}{4}) - (\dfrac{18L^2}{16})}[/tex]

[tex]\mathbf{A = (\dfrac{9L^2}{4}) - (\dfrac{9L^2}{8})}[/tex]

[tex]\mathbf{A = \dfrac{18L^2 - 9L^2}{8}}[/tex]

[tex]\mathbf{A = \dfrac{ 9L^2}{8}}[/tex]

Therefore, we can conclude that the largest possible area for the vegetable garden is [tex]\mathbf{ \dfrac{ 9L^2}{8}}[/tex]

Learn more about the area of a rectangle here:

https://brainly.com/question/9655204?referrer=searchResults

MrRoyal

The perimeter of a garden is the sum of its side lengths.

The largest possible area of the garden is [tex]\mathbf{ \frac {9}{8}L^2}[/tex]

The perimeter is given as:

[tex]\mathbf{P = 3L}[/tex]

Let the dimensions of the garden be x and y.

So, we have:

[tex]\mathbf{2x + y = 3L}[/tex]

Make y the subject

[tex]\mathbf{y = 3L - 2x}[/tex]

The area of the garden is:

[tex]\mathbf{A = xy}[/tex]

Substitute [tex]\mathbf{y = 3L - 2x}[/tex] in [tex]\mathbf{A = xy}[/tex]

[tex]\mathbf{A = x(3L - 2x)}[/tex]

Expand

[tex]\mathbf{A = 3xL - 2x^2}[/tex]

Differentiate

[tex]\mathbf{A' = 3L - 4x}[/tex]

Set to 0

[tex]\mathbf{3L - 4x = 0}[/tex]

Add 4x to both sides

[tex]\mathbf{4x = 3L}[/tex]

Solve for x

[tex]\mathbf{x = \frac 34L}[/tex]

Substitute [tex]\mathbf{x = \frac 34L}[/tex] in [tex]\mathbf{A = x(3L - 2x)}[/tex]

[tex]\mathbf{A = \frac 34L(3L - 2 \times \frac 34 L)}[/tex]

Expand

[tex]\mathbf{A = \frac 94L^2 - \frac 98 L^2}[/tex]

Take LCM

[tex]\mathbf{A = \frac {18L^2 - 9L^2}{8}}[/tex]

[tex]\mathbf{A = \frac {9L^2}{8}}[/tex]

[tex]\mathbf{A = \frac {9}{8}L^2}[/tex]

Hence, the largest possible area of the garden is [tex]\mathbf{ \frac {9}{8}L^2}[/tex]

Read more about maximum areas at:

https://brainly.com/question/11906003

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