fk378 said:Homework Statement
A total of x feet of fencing is to form 3 sides of a level rectangular yard. What is the maximum possible area of the yard, in terms of x?
The Attempt at a Solution
So far I have 2 sides of the yard is denoted by y, and one side is denoted by z.
Then, 2y+z=x
We want to maximize the Area=yz
Now, do I substitute one of the variables in the Area equation to get the answer? Then I know I need to differentiate, but with respect to what? I'm confused here.
fk378 said:Ok, so I have
A=yz=y(x-2y)=xy-2y^2
A'=y-4y
We want to find the critical points so set A'=0
A'=y-4y=0
y=4y
...that doesn't make sense...?
The maximum area of a rectangle with a given amount of fencing can be determined using the formula A = (x/4)^2, where x is the length of the fencing. This formula is derived from the fact that a rectangle has two equal sides and two unequal sides, and the maximum area is achieved when the two unequal sides are equal to half of the total fencing length.
Yes, a square will always result in the maximum area for a given amount of fencing. This is because a square has four equal sides, making it the most efficient shape for maximizing area with a limited amount of fencing.
The length of the fencing directly affects the maximum area of the rectangle. The longer the fencing, the larger the maximum area will be. This is because a longer fencing length allows for longer side lengths, which in turn creates a larger area.
No, the maximum area can only be achieved with specific dimensions of the rectangle. As mentioned before, the two unequal sides of the rectangle must be equal to half of the fencing length. This means that the length and width of the rectangle must be equal, resulting in a square.
The length and width of the rectangle can be found by dividing the fencing length by 4, as both sides of the rectangle will be equal to half of the total fencing length. This will result in a square shape, which will have the maximum area for the given amount of fencing.