Module 2: Lesson 3

 

Self-Check 3
  1.  Question 18 on page 101 of the textbook
    1.  
      1. This question involves just a substitution on the right side, so the question can be solved algebraically.

         


        The minimum area of canvas required for the walls, correct to the nearest square metre, is 130 m2.
      2. In order to solve the question graphically, certain information needs to be substituted into the formula:

         


        There are two methods that can be used to solve this equation.

        Method 1: Use a System of Two Functions

        Graph the following functions, and determine the x-value of the intersection point of the two functions.

         


         
        This shows graphs two functions. The first is surface area equals pi times radius times begin square root 36 plus radius squared end square root. The second function is surface area equals 16. The intersection point of the two functions is labelled at (6, 160).

        The radius of the base would be 6 m.

        Method 2: Use a Single Function

        Rearrange the equation  so that it equals zero.

         

         

        Graph the function , and then determine the x-intercept of the graph.

         
        This is the graph of the function surface area equals pi times radius times begin square root 36 plus radius squared end square root subtract 16. The x-intercept is labelled at (6, 0).

        The radius of the base would be 6 m.


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