The 'Dragon's Den' is a reality television show where entrepreneurs pitch their business ideas in the hopes of securing an investment from the team of investors also known as "The Dragons". (With the information in the attachment can you find the answers for these questions) Question 1 (1 point) Choose a selling price between $5-$10. Suppose on an average month you sell 500 items. For every $0.15 increase in your price you sell 10 less items. a) Find a function for Revenue, with respect to the number of price increases. Your function will be = (number of products sold) x (price). It is important to incorporate that for every $0.15 in your price you sell 10 less items. [Hint: First find a function for the number products sold, and a function for the price of the item.] p0 = $8 q0 = 500 p=0.15 q=10 The slope is: m= q 10 200 = = p 0.15 3 Now, the equation is follow: qq 0=m ( p p0 ) q500= 200 ( p8 ) 3 3 q1500=p+ 8 3 q=1508 p q= 1508 p 3 3 b) Provide a graph for your Revenue Function. R= pq= p p ) ( 1508 3 3 2 R= 1508 p p 3 3 c) Determine the roots of and interpret them in the context of the question. p p )=0 ( 1508 3 3 For p = 0 and 1508 p =0 3 3 Clearing p we have p 1508 = 3 3 p=1508 This values mean that for p=0 and p=$1508, we have a null revenue d) Determine the maximum or minimum values of and interpret them in the context of the question. The maximum value is given for the vertex of the parabola, in p = 1508/2 = 754 And 2 ( 754 ) 1508 R ( 754 )= ( 754 ) =189,505.33 3 3 This values mean that for 754 units sold we have a maximum revenue of $189,505.33 e) Determine the domain and range of and interpret them in the context of the question The domain is Dom R ( x ) ( 0, ) This interval corresponds to the price, which can start in 0 an so on Range R ( x ) (189,505.33 ,) This interval mean that there are an maximum value of 189,505.33 but can include also negative numbers to - f) Now create a revenue function with respect to the number of items sold and the new selling price. 2 R= If 1508 p p 3 3 we replace p Remember that 3 q=1508 p The we clear q as p=1508+3 q Now ( 1508+3 q ) 1508 R= ( 1508+3 q ) 3 3 Grouping 2 2 R(q)=3 q 1508 q Question 2 (1 point) Using your Cost and Revenue functions, find the profit function, . The cost per item is $2 So, C(q) = 2q Now, the profit function is P ( q )=R ( q ) - C ( q )=3 q21508 q2 q Finally P ( q )=3 q 21510 q The 'Dragon's Den' is a reality television show where entrepreneurs pitch their business ideas in the hopes of securing an investment from the team of investors also known as "The Dragons". (With the information in the attachment can you find the answers for these questions) Question 1 (1 point) Choose a selling price between $5-$10. Suppose on an average month you sell 500 items. For every $0.15 increase in your price you sell 10 less items. a) Find a function for Revenue, with respect to the number of price increases. Your function will be = (number of products sold) x (price). It is important to incorporate that for every $0.15 in your price you sell 10 less items. [Hint: First find a function for the number products sold, and a function for the price of the item.] p0 = $8 q0 = 500 p=0.15 q=10 The slope is: m= q 10 200 = = p 0.15 3 Now, the equation is follow: qq 0=m ( p p0 ) q500= 200 ( p8 ) 3 3 q1500=p+ 8 3 q=1508 p q= 1508 p 3 3 b) Provide a graph for your Revenue Function. R= pq= p p ) ( 1508 3 3 2 R= 1508 p p 3 3 c) Determine the roots of and interpret them in the context of the question. p p )=0 ( 1508 3 3 For p = 0 and 1508 p =0 3 3 Clearing p we have p 1508 = 3 3 p=1508 This values mean that for p=0 and p=$1508, we have a null revenue d) Determine the maximum or minimum values of and interpret them in the context of the question. The maximum value is given for the vertex of the parabola, in p = 1508/2 = 754 And 2 ( 754 ) 1508 R ( 754 )= ( 754 ) =189,505.33 3 3 This values mean that for 754 units sold we have a maximum revenue of $189,505.33 e) Determine the domain and range of and interpret them in the context of the question The domain is Dom R ( x ) ( 0, ) This interval corresponds to the price, which can start in 0 an so on Range R ( x ) (189,505.33 ,) This interval mean that there are an maximum value of 189,505.33 but can include also negative numbers to - f) Now create a revenue function with respect to the number of items sold and the new selling price. 2 R= If 1508 p p 3 3 we replace p Remember that 3 q=1508 p The we clear q as p=1508+3 q Now ( 1508+3 q ) 1508 R= ( 1508+3 q ) 3 3 Grouping 2 2 R(q)=3 q 1508 q Question 2 (1 point) Using your Cost and Revenue functions, find the profit function, . The cost per item is $2 So, C(q) = 2q Now, the profit function is P ( q )=R ( q ) - C ( q )=3 q21508 q2 q Finally P ( q )=3 q 21510