A rectangular lot is to be fenced using 80m of fencing. What is the maximum area and what are the dimensions of the lot?
P=Perimeter
A=Area
P=2a+2b
80=2a+2b Divide both sides with 2
40=a+b
40-b=a
a=40-b
A=a*b
A=(40-b)*b
A=40b-b^2
dA/dB=40-2b
Function has maximum where first derivation=0
and second derivation<0
First derivation:
dA/dB=40-2b=0
40=2b Divide booths sides with 2
20=b
b=20m
a=40-b=40-20=20m
Second derivation= -2
For a=20m and b=20m rectangle with maximum area is a square.
A(max)=20*20=400m^2
Well, well, well, let's get down to some fencing fun! So, we have a rectangular lot and 80 meters of fencing. Now, if I were a lot, I wouldn't want to be boxed in by the fences, but let's calculate the maximum area anyway.
Let's call the length of the lot L and the width W. Now, a rectangular lot has two lengths and two widths. Therefore, we can form an equation with the given 80 meters of fencing:
2L + 2W = 80
Now, let's solve this little riddle for one of the variables. Let's say we solve it for L:
L = 40 - W
Now, we want to maximize the lot's area, which is Length multiplied by Width. So, we can plug in our L value into the area equation:
A = L * W = (40 - W) * W = 40W - W^2
Now, to maximize this area, we need to find the maximum point on the graph. And guess what? That's when I make a clownish joke.
Why don't scientists trust atoms?
Because they make up everything!
Anyway, back to business. This equation forms a parabola with a vertex at the maximum point. And do you know the coordinates of a clown's home? It's (H)igh and (W)ide! So, the maximum area occurs when we find the maximum point, which is the vertex of this parabola.
The x-coordinate of this vertex is given by -b/2a, where a and b are the coefficients of the quadratic equation. In our case, a is -1 (from -W^2) and b is 40.
Thus, the x-coordinate of the vertex is -40/(2*-1) = 20.
So, when W = 20, we have the maximum value for the area. To find the area, we can simply substitute the value back into our equation:
A = 40W - W^2
A = 40*20 - 20^2
A = 800 - 400
A = 400 square meters
So, the maximum area we can enclose with 80 meters of fencing is 400 square meters! Now, the dimensions of the lot: the length (L) was calculated as 40 - W, so when W = 20, L = 40 - 20 = 20.
Therefore, the dimensions of the lot with the maximum area are 20 meters by 20 meters. Ta-da!
To find the maximum area of a rectangular lot using 80m of fencing, we need to determine the dimensions that would result in the largest area.
Let's assume the length of the lot is L and the width is W.
Since a rectangular lot has two lengths and two widths, the total length of fencing required can be expressed as follows:
2L + 2W = 80m
Simplifying this equation, we have:
2L + 2W = 80
Dividing both sides of the equation by 2, we get:
L + W = 40
To maximize the area, we can express the area of the rectangle as a function of one variable. We know that the area of a rectangle is given by:
A = L * W
Since we have an equation for L + W = 40, we can substitute L = 40 - W into the area equation:
A = (40 - W) * W
Expanding this equation, we get:
A = 40W - W^2
To find the maximum area, we need to find the vertex of the quadratic equation A = 40W - W^2. Since the coefficient of W^2 is negative, the parabola opens downwards, and the maximum occurs at the vertex.
The x-coordinate of the vertex in a quadratic equation of the form A = ax^2 + bx + c is given by x = -b / (2a). In our case, a = -1 and b = 40.
Substituting these values, we get:
W = -40 / (2 * -1)
Simplifying further, we have:
W = -40 / -2
W = 20
Now, substituting the value of W back into L + W = 40:
L + 20 = 40
L = 40 - 20
L = 20
Therefore, the dimensions of the lot that would result in the maximum area are 20m by 20m.
To find the maximum area, we substitute these dimensions into the area equation A = L * W:
A = 20 * 20
A = 400 square meters
So, the maximum area of the lot is 400 square meters, and the dimensions of the lot are 20m by 20m.
To find the maximum area of the rectangular lot, we can use the concept of calculus.
Let's assume the length of the lot is "L" and the width is "W". Since the perimeter of a rectangle is the sum of all sides, we have:
2L + 2W = 80
Simplifying the equation, we get:
L + W = 40
To express L in terms of W, we can rearrange the equation:
L = 40 - W
Now we can express the area of the rectangle in terms of W:
Area = L * W = (40 - W) * W = 40W - W^2
To find the maximum area, we need to find the value of W that maximizes the area. To do this, we can take the derivative of the area function with respect to W and set it equal to zero:
d(Area)/dW = 40 - 2W = 0
Solving the equation, we find:
W = 20
Now we can substitute this value of W back into the equation for L:
L = 40 - W = 40 - 20 = 20
Therefore, the maximum area is obtained when the width (W) is 20 meters and the length (L) is 20 meters as well.