a)
To form: a system of inequalities performed by Sean to paint sheds and play houses.
a)
Answer to Problem 25PPS
(a) Inequalities are:
(c) 25 sheds and 20 play houses
(d) So, maximum profit= $1250
Explanation of Solution
Given Information
Let the numbers of sheds be ‘ x’ and the number of play houses be ‘ y’ .
Time required to paint 1 shed = 2.5 days and the required to paint 1 playhouse = 2 days
But, Sean has a total of 20 days to paint
So,
⇒
⇒
Converting it in intercept form,
⇒
But, it is also given that, in total there are 45 structures,
So,
Converting it in intercept form,
Hence, our final inequalities are:
b)
To draw: a graph showing feasible region and its coordinates.
b)
Explanation of Solution
Given Information
Let the numbers of sheds be ‘ x’ and the number of play houses be ‘ y’ .
Time required to paint 1 shed = 2.5 days and the required to paint 1 playhouse = 2 days
But, Sean has a total of 20 days to paint
Now we will plot all these inequality in graph
Co-ordinates in the feasible region:
A = (0, 40)
B = (45, 0)
P = (25, 20)
c)
To find: how many sheds and play houses must be painted
c)
Answer to Problem 25PPS
(c) 25 sheds and 20 play houses
Explanation of Solution
Given Information
Let the numbers of sheds be ‘ x’ and the number of play houses be ‘ y’ .
Time required to paint 1 shed = 2.5 days and the required to paint 1 playhouse = 2 days
But, Sean has a total of 20 days to paint
Sean should paint 25 sheds and 20 play houses, in order to maximize the profit.
d)
To find: how many sheds and play houses must be painted for maximum profit
d)
Answer to Problem 25PPS
(d) So, maximum profit= $1250
Explanation of Solution
Given Information
Let the numbers of sheds be ‘ x’ and the number of play houses be ‘ y’ .
Time required to paint 1 shed = 2.5 days and the required to paint 1 playhouse = 2 days
But, Sean has a total of 20 days to paint
Maximum profit
Profit per shed = $ 26
Profit per play house = $ 30
So, maximum profit
Chapter 3 Solutions
Glencoe Algebra 2 Student Edition C2014
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