Chapter 7 - Algebra: Graphs, Functions, and Linear Systems - 7.3 Systems of Linear Equations in Two Variables - Exercise Set 7.3 - Page 445: 59

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(a) The cost function of the company can be written as a sum of the fixed cost of the company and the cost to produce each performance. If \[x\] is the number of performances produced and sold, then the cost function of the company is written as:\[C\left( x \right)=30,000+2500x\]. Thus, the cost function of the company is\[C\left( x \right)=30,000+2500x\]. (b) The revenue function of the company can be written as a product of price of a single performance and the total number of performances produced and sold. If \[x\] is the number of performances produced and sold, then the revenue function of the company is written as: \[R\left( x \right)=3125x\]. Thus, the revenue function of the company is\[R\left( x \right)=3125x\]. (c) Consider the formula for cost function calculated in part (a) and revenue function calculated in part (b). The no. of performances for which the cost function and revenue function are equal is known as the break-even point. The break-even point is found by the intersection of the two functions. \[\begin{align} & 3125x=30,000+2500x \\ & 625x=30,000 \\ & x=\frac{30,000}{625} \\ & x=48 \end{align}\] Thus, the break-even point of the company is \[48\]performance. Back – substitution 48 for x in either of the system’s equations Now, \[C(x)=30,000+2500x\text{ and }R(x)=3125x\] Now, \[\begin{align} & R(48)=3125(48) \\ & =150,000 \\ \end{align}\] The break – even point is \[(48,150000)\]. This means that the company will break even if it produces and sells 48 performance. At this level, the money coming in is equal to the money going out:\[\$150,000\]