I cant figure out this homework question! I need to use optimisation to find the answer but I cant work it out :(
There is a rectangular garden which is in need of fencing. Three sides of the 30m wide garden are already fenced. You own an additional 20m of suitable fencing and a 1m wide suitable gate. Without moving any of the existing fencing, what is the maximum area you can fence off using straight sections of fence and a corner of the existing fence to make a rectangle.
I believe you would have to input the rule: Area = L X W, somehow. But I don't know how the working would go!
To solve this problem using optimization, we need to find the maximum area that can be fenced off using the given materials.
Let's start by visualizing the problem. We have a rectangular garden with three sides already fenced and an additional 20m of fencing plus a 1m wide gate available. We want to find the largest possible area that we can enclose within this boundary.
Since we are working with rectangular shapes, we can use the formula for the area of a rectangle: Area = Length × Width.
Let's assume that the side of the garden that needs fencing is the length of the rectangle we want to maximize the area for, and the other two sides of the garden form the width (since they are already fenced and cannot be changed).
Let's denote the length of the fence we need to add as L. The width of the fenced-off area will be W, which is equal to the width of the garden. The length of the garden is already given as 30m.
To maximize the area, we need to maximize the product of L and W, which is the same as maximizing L since W is a constant (30m).
Now, since we only have 20m of fencing available (plus the 1m gate), we need to consider how to use this fencing efficiently.
To enclose the rectangular area, we need two lengths of fence, each equal to L, and one width of fence, equal to W. Additionally, we need one corner formed by the existing fence.
The total length of the fence we use will be: 2L + W + 1 (one corner and one gate).
Given that we have 20m of fencing available, we can express this as an equation:
2L + W + 1 = 20
Rearranging the equation, we have:
2L + W = 19
Since we need to express the area in terms of a single variable (L), we can substitute W with its value (30m) because it is constant. Therefore, we have:
2L + 30 = 19
Simplifying the equation, we get:
2L = 19 - 30
2L = -11
L = -11/2
However, negative lengths do not make sense in this context, so we can eliminate this result.
Therefore, we conclude that it is not possible to maximize the area by enclosing a rectangular region using the given materials.