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

The largest possible area of the vegetable garden is $\frac 92L^2$

How to determine the largest possible area

Let the side lengths of the fence be x and y.

So, the perimeter (P) is:

$P = x + 2y$

The perimeter is given as 6L.

So, we have:

$x + 2y = 6L$

Make x the subject

$x = 6L-2y$

The area of the field is:

$A = xy$

Substitute $x = 6L-2y$

$A = (6L - 2y) * y$

Expand

$A = 6Ly - 2y^2$

Differentiate

$A' = 6L - 4y$

Set to 0

$6L - 4y = 0$

Collect like terms

$4y = 6L$

Make y the subject

$y = \frac 32L$

Recall that:

$A = (6L - 2y) * y$

So, we have:

$A = (6L - 2 * \frac 32L ) * \frac 32L$

Simplify

$A = (6L - 3L ) * \frac 32L$

Evaluate the difference

$A = 3L * \frac 32L$

Evaluate the product

$A = \frac 92L^2$

Hence, the largest possible area of the vegetable garden is $\frac 92L^2$

Read more about areas at:

https://brainly.com/question/24487155