§ 0.6 Functions and Graphs in Application

Section Outline Geometric Problems Cost, Revenue, and Profit Surface Area Functions and Graphs

Geometric Problems EXAMPLE SOLUTION (Fencing a Rectangular Corral) Consider a rectangular corral with two partitions, as shown below. Assign letters to the outside dimensions of the corral. Write an equation expressing the fact that the corral has a total area of 2500 square feet. Write an expression for the amount of fencing needed to construct the corral (including both partitions). SOLUTION First we will assign letters to represent the dimensions of the corral. y x x x x y

Geometric Problems CONTINUED Now we write an equation expressing the fact that the corral has a total area of 2500 square feet. Since the corral is a rectangle with outside dimensions x and y, the area of the corral is represented by: Now we write an expression for the amount of fencing needed to construct the corral (including both partitions). To determine how much fencing will be needed, we add together the lengths of all the sides of the corral (including the partitions). This is represented by:

Cost Problems EXAMPLE SOLUTION (Cost of Fencing) Consider the corral of the last example. Suppose the fencing for the boundary of the corral costs $10 per foot and the fencing for the inner partitions costs $8 per foot. Write an expression for the total cost of the fencing. SOLUTION This is the diagram we drew to represent the corral. y x x x x y Since the boundary of the fence is represented by the red part of the diagram, the length of fencing for this portion of the corral is x + x + y + y = 2x + 2y. Therefore the cost of fencing the boundary of the fence is (2x + 2y)(cost of boundary fencing per foot) = (2x + 2y)(10) = 20x + 20y.

Cost Problems CONTINUED Since the inner partitions of the fence are represented by the blue part of the diagram, the length of fencing for this portion of the corral is x + x = 2x. Therefore the cost of fencing the inner partitions of the fence is (2x)(cost of inner partition fencing per foot) = (2x)(8) = 16x. Therefore, an expression for the total cost of the fencing is: (cost of boundary fencing) + (cost of inner partition fencing) (20x + 20y) + (16x) 36x + 20y

Surface Area EXAMPLE SOLUTION Assign letters to the dimensions of the geometric box and then determine an expression representing the surface area of the box. SOLUTION First we assign letters to represent the dimensions of the box. z y x

Surface Area CONTINUED z y x Now we determine an expression for the surface area of the box. Note, the box has 5 sides which we will call Left (L), Right (R), Front (F), Back (B), and Bottom (Bo). We will find the area of each side, one at a time, and then add them all up. L: yz R: yz F: xz B: xz Bo: xy Therefore, an expression that represents the surface area of the box is: yz + yz + xz + xz + xy = 2yz +2xz +xy.

Cost, Revenue, & Profit EXAMPLE SOLUTION (Cost, Revenue, and Profit) An average sale at a small florist shop is $21, so the shop’s weekly revenue function is R(x) = 21x where x is the number of sales in 1 week. The corresponding weekly cost is C(x) = 9x + 800 dollars. What is the florist shop’s weekly profit function? How much profit is made when sales are at 120 per week? If the profit is $1000 for a week, what is the revenue for the week? SOLUTION (a) Since Profit = Revenue – Cost, the profit function, P(x), would be: P(x) = R(x) – C(x) P(x) = 21x – (9x + 800) P(x) = 21x – 9x - 800 P(x) = 12x - 800

Cost, Revenue, & Profit CONTINUED (b) Since x represents the number of sales in one week, to determine how much profit is made when sales are at 120 per week, we will replace x with 120 in the profit function and then evaluate. P(120) = 12(120) - 800 P(120) = 1,440 - 800 P(120) = 640 Therefore, when sales are at 120 per week, profit is $640 for that week. (c) To determine the revenue for the week when the profit is $1000 for that week, we use an equation that contains profit, namely the profit function: P(x) = 12x - 800 Now we replace P(x) with 1000 and solve for x. 1000 = 12x - 800 1800 = 12x 150 = x Therefore x, the number of units sold in a week, is 150 when profit is $1000.

Cost, Revenue, & Profit CONTINUED Now, to determine the corresponding revenue, we replace x with 150 in the revenue function. R(x) = 21x R(150) = 21(150) R(150) = 3,150 Therefore, when profit is $1000 in a week, the corresponding revenue is $3,150. NOTE: In order to determine the desired revenue value in part (c), we needed to solve for R(x). But in order to do that, we needed to have a value for x to plug into the R(x) function. In order to acquire that value for x, we needed to use the given information – profit is $1000.

Functions & Graphs EXAMPLE The function f (r) gives the cost (in cents) of constructing a 100-cubic-inch cylinder of radius r inches. The graph of f (r) is given. What is the cost of constructing a cylinder of radius 6 inches? Interpret the fact that the point (3, 162) is on the graph of the function. Interpret the fact that the point (3, 162) is the lowest point on the graph of the function. What does this say in terms of cost versus radius?

Functions & Graphs CONTINUED SOLUTION To determine the cost of constructing a cylinder of radius 6 inches, we look on the graph where r = 6. The corresponding y value will be the cost we are seeking. The red arrow is emphasizing the point in which we are interested. The y value of that point is 270. Therefore, the cost of constructing a cylinder of radius 6 inches is 270 cents or $2.70.

Functions & Graphs CONTINUED (b) The fact that the point (3, 162) is on the graph tells us that the cost to make 100-cubic-inch cylinders with a radius as small as 3 inches is 162 cents or $1.62. (c) The fact that the point (3, 162) is the lowest point on the graph tells us that the least expensive 100-cubic-inch cylinder that can be made is a 3 inch cylinder at a cost of $1.62. Therefore, the 3 inch cylinder is the most cost-effective one that is offered.