Question

You have 6L 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

Answer 1

Answer:

A(max) = (9/2)*L² ft²

Dimensions:

x = 3*L feet

y  = (3/2)*L   ft

Step-by-step explanation:

Let call  "x"  and  " y " sides of the rectangle. The side x is parallel to the wall of the house then

Area of the rectangle is  

A(r) = x*y

And total length of fence available is 6*L f , and we will use the wall as one x side then, perimeter of the rectangle which is 2x  +  2y   becomes x + 2*y

Then

6*L  =  x + 2* y    ⇒ y  = ( 6*L - x ) /2

And the area as function of x is

A(x)  =   x* ( 6*L - x )/2

A(x)  = ( 6*L*x  - x² ) /2

Taking derivatives on both sides of the equation we get:

A´(x)  =  1/2 ( 6*L - 2*x )

A´(x)  =  0      ⇒      1/2(  6*L  - 2*x ) = 0

6*L - 2*x  = 0

-2*x  = - 6*L

x  =  3*L feet

And

y  = ( 6*L - x ) /2   ⇒  y  = ( 6*L - 3*L )/ 2

y = ( 3/2)*L feet

And area maximum is:

A(max)  =  3*L * 3/2*L

A(max)  = (9/2)*L² f²

Answer 2

Answer:

Maximum Area of A = $\frac{9L^2}{2} \: ft^{2}$

Where the dimensions are:

x = 3L feet

y  = $\frac{3L}{2} \: ft$

Step-by-step explanation:

Let the dimensions of the rectangular vegetable garden be: a and b.

Where the side a is parallel to the wall of the house.

The total length of fencing =6L ft

Perimeter of the rectangle=2(Length+Breadth)

However, since the wall is parallel to side a, there will be only one side of length a.

Therefore:

Perimeter of the Fence

6L=a+2b

⇒$b=\frac{6L-a}{2}$

Area of a rectangle=Length X Breadth

Then the area of the Rectangle=ab

The area of the garden in terms of a is then given as:

$A(a)=\frac{a(6L-a)}{2}$

$A(a)=\frac{6aL-a^2}{2}$

To find the maximum area(the vertex of the function), we need to determine the point at which the derivative of A(a) is zero.

Taking derivatives:

$A^{'}(a)=\frac{6L-2a}{2}$

Setting $A^{'}(a)=0$

$\frac{6L-2a)}{2}=0$

-2a=-6L

a= 3L ft.

Recall: $b=\frac{6L-a}{2}$

$b=\frac{6L-3L}{2}$

$b=\frac{3L}{2} ft$

The maximum area therefore is:

$Area=\frac{3L}{2} X 3L ft^{2}$

$=\frac{9L^2}{2} ft^{2}$