Hi everyone
I am lost on how to do this math problem:
A. You have 180 feet of fencing with which to construct a rectangular pen. What is the maximum area you can enclose?
B. Going back to part A, you decide to build your rectangular pen next to a large barn. (Assume the barn is more than 180 feet long, or at least long enough to be used as one side of your pen.) Now what is the maximum area you can enclose?
What I know is an area of a rectangle is (base)(height) but I'm lost on how to do this problem exactly.
Any help is greatly appreciated!
it would be nice to know how much derivative calculus you know.
a) area=LW
Perimter=180=2L+2W
or W=90-L
Area=L(90-L)=90L-L^2
Ok: non calculus approach. That equation is a parabola, with roots at L=0, or L=90. So the max must be halfway, or L=45. Then W=45. Max area=45^2
Now calculus approach:
Area=90L-L^2
dArea/dL= 90-2L=0 or L=45, and W-45
b. Area=LW
180=L+2W
Area=(18O-2W)W
Area= 180W-2W^2
zeroes at W=90, W=0, so max is again at W=45, L=90
calculus approach:
Area=(180W-2W^2
dArea/dW=180-4W=0 W=45, L=90
For a given perimeter, the largest area a rectangle can have is when the rectangle is a square, so
a) side = 45 ft
area = 45^2 or 2025 ft^2
b) now we need only one length, call it y ft
let the width be x ft
2x + y = 180
y = 180-2x
area = xy = x(180-2x)
= -2x^2 + 180x
this is a quadratic function whose graph opens downwards, thus we have a maximum area
the vertex will give us that.
The x of the vertex is -b/(2a)
= -180/-4 = 45
so the width is 45 ft and the
length = 90 ft
for a maximum area of 4050 ft^2
Thanks for the help, I just started this math course and I'm pretty sure I don't know much derivative calculus yet!
To solve these math problems, we can use the concept of optimization. We want to find the maximum area that can be enclosed using a given amount of fencing.
A. To find the maximum area of a rectangular pen using 180 feet of fencing, we can start by considering the properties of a rectangle.
Let's assume the length of the rectangle is L and the width is W. Since the pen is rectangular, we have two equal lengths and two equal widths.
We know that the perimeter of the rectangle is equal to the sum of all four sides. In this case, the perimeter is 180 feet, so we can write an equation:
2L + 2W = 180
Simplifying the equation, we have:
L + W = 90
To determine the maximum area, we need to express the area (A) in terms of one variable. We can rewrite the equation above as:
L = 90 - W
Now, we can express the area of the rectangle in terms of a single variable:
A = L × W = (90 - W) × W = 90W - W^2
To find the maximum area, we can take the derivative of the area equation with respect to W, set it equal to 0, and solve for W. Rearranging the equation:
dA/dW = 90 - 2W = 0
Solving this equation, we find W = 45. Substituting this value back into the equation for the area, we get:
A = 90(45) - (45)^2 = 2025 square feet.
Therefore, using 180 feet of fencing, the maximum area that can be enclosed by the rectangular pen is 2025 square feet.
B. Now, let's consider the scenario where you want to build the rectangular pen next to a large barn.
Assuming the barn serves as one side of the pen, we only need to use fencing for the remaining three sides. In this case, we have two widths (W) and one length (L), so the equation for the perimeter becomes:
2W + L = 180
Rearranging the equation, we get:
L = 180 - 2W
Using the same approach as before, we can express the area (A) in terms of W:
A = L × W = (180 - 2W) × W = 180W - 2W^2
Taking the derivative of the area equation with respect to W and setting it equal to 0:
dA/dW = 180 - 4W = 0
Solving for W, we find W = 45. Substituting this value back into the equation for the area, we get:
A = 180(45) - 2(45)^2 = 4050 square feet.
Therefore, when building the rectangular pen next to a large barn, using 180 feet of fencing, the maximum area that can be enclosed is 4050 square feet.
I hope this explanation helps! Let me know if you have any further questions.