Final answer:
To find the dimensions of the rectangle with the largest area enclosed by 3300 meters of fencing, we express the perimeter in terms of width and length, and then use calculus to find the maximum area. By solving equations and taking derivatives, we find that the dimensions for the largest area rectangle are 825 meters by 825 meters.
Step-by-step explanation:
The question asks to find the dimensions of the rectangle with the largest area that can be enclosed on all sides by 3300 meters of fencing. We can solve this problem by using the formula for the area of a rectangle, which is given by the product of its length and width. Since the perimeter of the rectangle is 3300 meters, we can express the perimeter in terms of the width and length and then use calculus to find the dimensions that maximize the area.
Let's denote the width of the rectangle as w and the length as l. We are given that the perimeter (P) is equal to 3300 meters. Using the formula for the perimeter of a rectangle, we have:
P = 2w + 2l
Substituting the given value for P:
3300 = 2w + 2l
Because we are looking for the maximum area, we should express the area (A) in terms of a single variable. The formula for the area of a rectangle is:
A = lw
Substituting the expression for l from the equation for the perimeter:
A = w(3300 - 2w)
We can now find the width that maximizes the area by taking the derivative of A with respect to w, setting it equal to zero, and solving for w. Let's differentiate A:
dA/dw = 3300 - 4w
Setting dA/dw equal to zero and solving for w:
3300 - 4w = 0
4w = 3300
w = 825
Now that we have the value for w, we can substitute it back into the equation for the perimeter to find l:
3300 = 2(825) + 2l
3300 = 1650 + 2l
1650 = 2l
l = 825
Therefore, the rectangle with the largest area that can be enclosed on all sides by 3300 meters of fencing has dimensions of 825 meters by 825 meters.
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