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Learn how to model various problems using Quadratics like maximizing revenue and area. Practice solving for two numbers, revenue optimization, and area maximization with step-by-step examples.
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MPM2D Quadratic Word Problems (revisited)
Quadratic Word Problems • Many types of problems can be modelled with Quadratics • Trajectories of objects under the influence of gravity • Bridges and supports • Generating maximum revenue • Maximizing areas of enclosures • Solving for 2 numbers
Quadratic Word Problems • Each problem will require you to assign algebraic letters to things you solve for • Many times it will be a letter for each of the dependent and independent variables • Choose a letter that makes sense: R for revenue, P for profit, A for area, etc • Make “Let” statements: Let “x” represent one number and “w” (or whatever) the other number. (Don’t use y as we need it for the equations later.) • Reword the question in your own words
Quadratic Word Problems • Example: Two numbers have a difference of 12 and their product is a minimum. • Let x be one number and z be the other number. x – z = 12 x – 12 = z We know that xz is a minimum so it is the valley of a quadratic that opens up so an equation can be used: y = x (x – 12)
Quadratic Word Problems y = x (x – 12) = x2 – 12x Completing the square: y = x2 – 12x = x2 – 12x + 36 – 36 = (x – 6)2 – 36 The vertex is: (6, –36) So x = 6 is one of the numbers. z = x – 12 = 6 – 12 = – 6 The two numbers are 6 and – 6.
Another question • Find two numbers whose sum is 48 and whose product is a maximum.
Revenue • With Revenue: you let R be the revenue and the number of increases in the cost of the item be x or whatever. • Usually the question starts with the cost of an item selling a certain number and when the cost is raised, fewer items are sold. • The maximum (peak of the quadratic) shows when max revenue is generated (when the number of items and their cost yields the best revenue!!)
Revenue Question • The Enviro Club sells Sweatshirts to raise money. They sell 1200 shirts a year at $20 each. A survey shows that if the price were to be raised by $2, they would lose 60 sales that year. What selling price would yield the highest revenue? • Let R be the revenue, and x the number of cost increases.
Sweatshirt Sales • The cost of a sweatshirt is (20 + 2x) • The number sold is (1200 – 60x) • The Revenue, R, is found by multiplying the cost of the sweatshirts by the number sold. R = (20 + 2x)(1200 – 60x) • Find the maximum of this function.
Area • The maximum area of a rectangle, oval, triangle, circle or whatever may be useful for landscapers. • Area is a product of 2 dimensions (like base and height for a triangle). • Let L be one dimension and w be another for rectangles/squares. • Use b and h for triangles.
Area Question • Determine the maximum area of a triangle in square centimetres. The sum of its base and height must be 10 cm. • Let b be the base and h be the height (naturally!) h + b = 10 h = 10 – b Area = bh/2 A = 0.5(b)(10 – b)
Area q contd. A = 0.5(b)(10 – b) Find the maximum area and get b.