SOLUTION: Please help me I really need this thank you Answer the given problem. 1. It costs a bakery Php 15 to make a loaf of bread and Php 30 to make a pineapple pie. Production costs on

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Question 1188342: Please help me I really need this thank you
Answer the given problem.
1. It costs a bakery Php 15 to make a loaf of bread and Php 30 to make a pineapple pie. Production costs on these items cannot exceed Php 300. There must at least be 5 of these items.
a. Write a system of inequalities that shows the various combinations the bakery can produce these
two items. (Let x be the number of loaf bread and y be the number of pineapple pie.)
b. Graph.
c. Write 3 possible combinations of loaves of bread and pineapple pies.

Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
let x = the number of loaves of bread.
let y = the number of pineapple pies.

the cost is equal to 15x + 30y.

the cost cannot exceed 300.

the inequality for that is 15x + 30y <= 300.

there must be at least 5 items.

the inequality for that is x + y >= 5.

your constraint inequalities are:

x + y >= 5
15x + 30y <= 300

using the desmos.com calculator, you would graph the opposite of these inequalities.

the area on the graph that is not shaded is your region of feasibility.

any coordinate point in the feasible region satisfies the requirements of the problem.

two other inequalities that need to be satisfied are x >= 0 and y >= 0 since the number of each can't be negative.

your constraint inequalities are:

x + y >= 5
15x + 30y <= 300
x >= 0
y >= 0

you graph the opposite of these inequalities.

here's what the graph looks like.



i chose 3 points in the feasible region.

they are:

(3,7)
(5,4)
10,2)

each of those points will satisfy every one of the constraint inequalities because they are in the region of feasibility.

algebraically, the constraints they must satisfy are:

the minimum value of x and y is 0.

the maximum value of x and y is defined by 15x + 30y <= 300.

when x = 0, 15x + 30y <= 300 becomes 30y <= 300 which results in y <= 10.
when y = 0, 15x + 30y <= 300 becomes 15x <= 300 which results in x <= 20.

your requirements are that x and y must each be greater than or equal to 0 and x must be less than or equal to 20 and y must be less than or equal to 10.

all these points satisfy those requirements.

the point (3,7) has 0 <= x <= 20 and 0 <= y <= 10.
the point (5,4) has 0 <= x <= 20 and 0 <= y <= 10.
the point (10,2) has 0 <= x <= 20 and 0 <= y <= 10.

on the graph, they are all in the feasible region, which is the unshaded area of the graph.

the graphing calculator used can be found at https://www.desmos.com/calculator

as an example:

when the point is (3,7), .....

x + y >= 5 becomes 3 + 7 >= 5 which becomes 10 >= 5 which is true.

15x + 30y <= 300 becomes 15 * 3 + 30 * 7 <= 300 which becomes 45 + 210 <= 300 which becomes 255 <= 300 which is also true.


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