A town is organizing a Fourth of July Parade. Solve the Problem by showing your work, solutions and your steps.

Question

Solve the Problem by showing your work, solutions and your steps.

A town is organizing a Fourth of July parade. There will be two sizes of floats in the parade, the first is 30ft and the second is 15ft.. A space of 10 ft will be left after each float.

A. Describe how the total length of the parade will be calculated.

B. The parade must be a least 150ft long, but less than 200ft long. What combinations of large and small floats are possible?

C. Large float costs $600 to operate. Small floats cost $300 to operate. The town has a budget of $2500 to operate the floats. How does this change your answers to part B? What combinations of large and small floats are possible?

Answer 1

A. The total length of the parade will be the sum of the lengths of the floats and the space between them. We can easily describe the length of the floats in this expression: 30X + 15Y, where X is the number of large floats and Y the number of small boats. It is necessary to multiply 30 to X and 15 to Y because X + Y would just give us the number of floats total. The total length must include the ten foot space between floats, to acknowledge this in our equation we can write 10(X + Y), to show there are ten feet for every float, and every float will be the sum of the large floats and the small floats. Our final equation is 30X + 15Y + 10(X + Y) = P, where P is total length of the parade. We can simplify this equation using 1. The Distributive Property; 30X +15Y + 10X + 10Y = P, and 2. Adding Common Terms; 40X + 25Y = P.

B. The question describes an inequality. The length P of the parade is at least (so it will be more than or equal to) 150 feet and less than 200 feet. If the length of the parade is P, then 150 < P and P < 200. We know that P = 40X + 25Y, so we can substitute; 150 < 40X + 25Y and 40X + 25Y < 200. As there are several solution this is most easily solved by graphing the inequalities and picking the coordinates within the graphed solution. The coordinates must be natural (positive & whole) numbers as there is no such thing as half of a float, a negative float, and it is stated that there are two types of floats in the parade so 0 floats of either type is not a possible solution. Solutions: (1,5), (1,6), (2,3), (2,4), (3,3), (3,2), (4,1).

C. To find the total cost of the floats we set up the expression 600X + 300Y, again we have to multiply to account for the number of floats costing 600$ for one large float, X amount of times. This total cost has to be less than or equal to the budget, so 600X + 300Y < 2,500. We can graph this equation to see which of our solutions from part B fall within this inequality. We can also solve this using a system of equations with the three equations we created; 1. 150 < 40X + 25Y, 2. 40X + 25Y < 200, and 3. 600X + 300Y < 2,500. Solutions: (3,3), (4,1). The two possible combinations are three large floats and three small floats, and 4 large floats and 1 small float.