OLD MCDONALD IS GOING TO PUT A PIG PEN UP AGAINST HIS BARN. HE HAS 45 FEET OF FENCING AND WILL ONLY NEED TO MAKE 3 ADDITIONAL SIDES, AS THE BARN ACTS AS THE FOURTH SIDE.. WHAT IS THE MAXIMUM AREA THAT THE PIG PEN CAN HAVE.
area = x * y
x + 2y = 45 ... x = 45 - 2y
substituting ... a = 45 y - 2 y^2
ymax = -b / 2a = -45 / (2 * -2) = 11.25
x = 22.5
a = 22.5 * 11.25 = 253.125
or,
by Calculus
short side --- x
single long side --- y
2x + y = 45
y = 45-2x
Area = A = xy = x(45-2x)
= 45x - 2x^2
dA/dx = 45 - 4x = 0 for a max of A
4x = 45
x = 45/4 , then y = 45-2(45/4) = 45/2
max area = xy = (45/4)(45/2)
= 2025/8
= appr 253.125 , same as Scott
To find the maximum area of the pig pen, we need to determine the dimensions that would result in the largest possible area.
Let's assume the barn is a rectangle with one side measuring x feet. This means the two shorter sides of the pen will also measure x feet (as they need to connect to the barn). The remaining side will measure 45 - 2x feet (since we need to subtract the total length of the sides connected to the barn from the total available fencing).
To find the area, we multiply the lengths of the two shorter sides:
Area = x * x = x^2
Now we can find the maximum area by finding the value of x that maximizes the area.
To do this, we can take the derivative of the area function and set it equal to zero:
d(Area)/dx = 2x = 0
Solving this equation gives us x = 0, but since we are dealing with a physical scenario, we know that the value of x cannot be zero. Therefore, there is no maximum area. The area increases indefinitely as x increases.
In this case, Old McDonald can make the area of the pig pen arbitrarily large as long as he has enough fencing (45 feet in this scenario).
To find the maximum area that the pig pen can have, we need to determine the dimensions of the pen that will use up the entire 45 feet of fencing most efficiently.
Let's assume the pen is in the shape of a rectangle, with the barn acting as one of the sides. The other three sides will be made using the 45 feet of fencing.
We can label the length of the pen as "L" and the width as "W." Since two sides are equal to the width and one side is equal to the length, we can set up the following equation based on the perimeter of the rectangle:
2W + L = 45
However, we need to express the area of the pen, which is equal to the length multiplied by the width. Let's solve the equation for L:
L = 45 - 2W
Now we can substitute the value of L into the equation for the area:
Area = L * W = (45 - 2W) * W = 45W - 2W^2
To find the maximum area, we need to find the value of W that maximizes the quadratic equation. One way to accomplish this is by graphing the equation and identifying the highest point on the curve. Another method is to use calculus by finding the critical points. However, let's use the first method of graphing for simplicity.
By graphing the equation Area = 45W - 2W^2 on a coordinate system, we can determine the maximum area by identifying the highest point on the curve.
Now, there are various ways to graph the equation such as using a graphing calculator or software like Excel, or you can manually create a table of values and plot the points. Once you have the graph, locate the highest point on the curve, and the corresponding value of W will give you the width of the pen that maximizes the area. Plug this value back into the equation to find the maximum area of the pig pen.
Please note that the graph should only show positive values for W since width cannot be negative in this context.