Related problems
When students are given problems to solve, the problems are grouped by the techniques that are used in solving them. When students learn how to solve linear equations, they expect problems that can be solved by this technique. When they learn about quadratic equations, they expect problems that lead to quadratic equations, and so on. But similar problems are often solved by very different methods, and the choice of method is often the most important step toward a solution.
An example of three similar problems.
1. Design and construct from poster board a cylinder that has a volume V = 800 cu. cm, and the radius of its base r = 5 cm.
Solution.
V = π*r2*h, where h is the height. Therefore h = V/(π*r2). Using a simple calculator, or paper and pencil, we get h = 10.2 cm. (Now the dimensions of the cylinder, r and h, are known, and it can be constructed.)
2. Design and construct from poster board a cylinder having volume V = 800 cu. cm, and height h = 8 cm.
Solution.
We need to find r. We know V = π*r2*h, where r is the radius of the base. So r = √(V/π/h), which is the same as √(V/(π*h)). Again, we need only a simple calculator to get r = 5.6 cm.
3. Design and construct from poster board a cylinder having volume V = 800 cu. cm, and total surface area A = 500 sq. cm.
Solution.
We need to find r and h.
V = π*r2*h, and A = 2*π*r*(r + h). Therefore h = V/(π*r2), and
2*π*r*(r + V/(π*r2)) = A. (1)
Multiplying equation (1) out, we have
2*π*r 2+ 2*V/r = A
Multiplying both sides by r, we have
2*π*r 3+ 2*V = A*r
Rewriting this, we have
2*π*r3 - A*r + 2*V = 0.
This is a cubic equation in r, and we know A = 500 sq. cm and V = 800 cu. cm.
2*π*r3 - 500*r + 1600 = 0
We need to know r and h. Once we find r, we can find h from the equation for V or the equation for A.
This cubic equation in r can be solved with a graphing calculator.
Y= 2πX3 - 500X + 1600
You can use CALC, which will yield two solutions, r = 6.21 cm, (and therefore h = 6.6 cm.), and r = 4.01 cm (and therefore h =15.9 cm). Now you can construct your cylinder!
Conclusion.
Three very similar problems dealing with a cylinder lead to first, second, and third degree equations. Such an arrangement of problems presents a more realistic rendering of applied mathematics, and this rendering is practically feasible when students use calculators for numerical computations.