Perimater=4L=2width+1length
or length=4L-2w
area=width(4L-2W)
darea/dw=(4L-2w) + w(-2)=0
width=L
length=2L
check my thinking
or length=4L-2w
area=width(4L-2W)
darea/dw=(4L-2w) + w(-2)=0
width=L
length=2L
check my thinking
1. Let's assume the length of the vegetable garden is L and the width is W.
2. The length of the garden is equal to the length of the wall of the house, which is L.
3. The remaining fencing is used to create the two other sides of the garden, so the overall fence length is: 2W + L.
4. Since the fence length is 4L, we can write the equation: 2W + L = 4L.
5. Solving the equation for W, we have: 2W = 3L, or W = (3L / 2).
6. Now we have the length L and the width W in terms of L.
7. The area of the rectangle is given by multiplying the length and width: A = L * W.
8. Substituting the values of L and W, we get: A = L * (3L / 2) = 3L^2 / 2.
So, the largest possible area for the vegetable garden is given by the equation A = 3L^2 / 2.