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!
Perhaps you would assume that because there is another 21m of fencing available, that the rectangle was 30m x 21m ??
I read this as you are to put the 21m(fence and gate)in one corner, so that the resulting rectange has two sides made up of the original fencing, and two sides totaling 21m. It would be an odd way to build a garden. In that case
area=LW
but L+W=21
area=L(21-L)
darea/dl=0=21-L + L(-1)
or L=21/2
W =21/2
max area= 21^2/4
I think the question means 3 sides of the rectangle are already fenced and you have 21m of extra fencing for the remaining side? :/
But I really have no idea..
To solve this problem using optimization, you need to find the dimensions of the rectangle that will maximize its area within the given constraints.
Let's start by visualizing the problem. You have a rectangular garden that is 30 meters wide and three sides are already fenced. You have 20 meters of additional fencing and a 1-meter wide gate. You need to use the given fencing and the corner of the existing fence to create a rectangle with the maximum possible area.
To find the optimal dimensions, let's consider the different possibilities. Since the width of the garden is fixed at 30 meters, we can use the additional fencing to enclose the remaining sides of the rectangle. Let's assume the length of the rectangle is x meters.
Since you have 20 meters of fencing and need to leave 1 meter for the gate, the lengths of the two sides will be (x - 1) meters and (20 - x) meters. The width will remain 30 meters.
Now, let's calculate the area of this rectangle using the formula Area = Length x Width. In this case, the area would be:
Area = (x - 1) * 30
We want to maximize this area, so we can consider it as a function of x. Let's call it f(x).
f(x) = (x - 1) * 30
Now, to find the optimal value of x, we can take the derivative of f(x) with respect to x and set it equal to zero. This will help us find the critical point where the area is maximized.
f'(x) = 30
Setting f'(x) = 0, we find that x has no effect on the derivative and therefore does not have an absolute maximum or minimum. However, since the domain of x is restricted to the interval (0, 20), we can see that the possible values for x are 1 to 19.
Now, we need to evaluate the function f(x) at the endpoints and critical points within the domain to determine the largest area.
f(1) = (1 - 1) * 30 = 0
f(19) = (19 - 1) * 30 = 540
Comparing the values, we can see that f(19) yields the maximum value for the area of the rectangle.
Therefore, the maximum area you can fence off using the given conditions is 540 square meters, which occurs when the length of the rectangle is 19 meters and the width is 30 meters.
I hope this explanation helps you understand the process of solving this optimization problem!