Example
A farmer plans to fence a rectangular grazing area along the length of a river with 225 yards of fence. Note that the farmer will not fence the length next to the river. The farmer wishes to make the grazing area as large as possible, so his herd has plenty of grass. Find the length and width of the grazing area and the area he can enclose.
a) Identify the unknowns and assign a variable to each.
b) Write an expression for the perimeter of the grazing area. Hint: P = 2L + 2W for the a rectangle with ALL FOUR SIDES.
c) Using the formula for area of a rectangle, substitute the expression for length so that the only length in terms of the width
d) Notice that the expression in Part (c) is a quadratic equation. What is the maximum area he can enclose? Answer in a complete sentence, including units. What is the width of this grazing area?
e) Substitute the width into the expression from the length of part (b) to find the length. Now give the dimensions of the grazing area.
Solution
a) Let L represent the length of the yard and W represent the width of the yard.
L: Length of the Grazing Area
W: Width of the Grazing Area
W: Width of the Grazing Area
b) Since we know that one length of the grazing area will not be needed since the river is along the length of the grazing area. The expression for the perimeter is L+2W, and since the total of the fence is 225, then we can have the following equation, which is we solve for L:
c) Since the grazing area is a rectangle, and the area of a rectangle is A+LW. Substitute the expression for L from above into the formula to obtain an expression for area in terms of width:
d) To find the maximum, we use the formula to find the x-coordinate of the vertex, where a and b are the coefficients of the quadratic equation. a=-2 and b=225, we find the x-coordinate of the vertex:
Remember, the x-coordinate represents the width of the yard with maximum area. Now, to find the maximum area, we either find the length and multiply or substitute into the formula for area:
The maximum area is 6328.125 square yards and the width is 56.25 yards.
e) The width of the grazing area is 56.25 yards. We also know that the area is 6328.125 yards. We can use the formula for length and substitute in the value for width:
The dimensions of the grazing area is 112.5 yards by 56.25 yards
EQUATION, TABLE, AND GRAPH
Reflection
This example is about what is the length and width of a grazing area that a farmer is trying to fence off for his livestock by a river. One length of the grazing area will not be needed since it's right next to the river so we need to find out what is the value of one of the lengths and the value of two of the widths. To solve this word problem we have to use a quadratic function. The equations that were use were L+2W=P, A=LW, and -b/2a. There equations are used to find the maximum area, length, width, and minimum length of the grazing area. The graph is displayed as a quadratic function as the problem follows quadratic functions. With the maximum being the top on that parabola. This word problem is a bit more difficult that the previous problems, but not so much that it would not take significantly longer to solve these problems. Overall the problem is a good start for quadratic equations before the more advanced formulas .