SOLUTION: The relationship between the number of calculators x that a company can sell per month and the price of each calculator p is given by x = 1700 - 100p. Find the price at which a c

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 Click here to see ALL problems on Polynomials-and-rational-expressions Question 342259: The relationship between the number of calculators x that a company can sell per month and the price of each calculator p is given by x = 1700 - 100p. Find the price at which a calculator should be sold to produce a monthly revenue of \$7000. (Hint: Revenue = xp.)Answer by jsmallt9(3438)   (Show Source): You can put this solution on YOUR website!With and we have a system of two equations with two variables. Since the second equation is of degree 2 (because of the x*p term), we will use the Substitution Method. (With second degree equations other methods are often either not possible or not practical.) Since we already have the first equation solved for x, we will use that and substitute into the second equation: To solve this we will start by simplifying: We can simplify the equation by dividing both sides by 100: Now we can factor (or use the Quadratic Formula): From the Zero Product property we know that: or solving these we get: p = 7 or p = 10 Now we find the x for each of these p's. We'll use the first equation and substitute in the value for p: For p = 7: For p = 10: So we have two solutions which generate \$7000 revenue: When the price is \$7, 1000 calculators will be sold. When the price is \$10, 700 calculators will be sold. If it is cheaper to build 700 calculators than it would be to build 1000 calculators, the greatest profit would come from making 700 calculators and selling them for \$10 each.