SOLUTION: The table shows the daily production level and profit for a business. Use the quadratic function {{{ y=ax^2+bx+c }}} to determine the number of units that should be produced eac

Algebra ->  Matrices-and-determiminant -> SOLUTION: The table shows the daily production level and profit for a business. Use the quadratic function {{{ y=ax^2+bx+c }}} to determine the number of units that should be produced eac      Log On


   

Question 1137649: The table shows the daily production level and profit for a business. Use the quadratic function +y=ax%5E2%2Bbx%2Bc+
to determine the number of units that should be produced each day for maximum profit. What is the maximum daily​ profit?
______________________________________________________________
x​ (Number of Units Produced​ Daily) | 30 | 50 | 100
______________________________________________________________
y​ (Daily Profit) | ​$5800 | $7320 | ​$4120


Each day [??] units should be produced to have a maximum daily profit of ​$[????].
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!


The table shows the daily production level and profit for a business. Use the quadratic function +y=ax%5E2%2Bbx%2Bc+
to determine the number of units that should be produced each day for maximum profit. What is the maximum daily​ profit?
______________________________________________________________
x​ (Number of Units Produced​ Daily) | 30 | 50+ | 100
______________________________________________________________
y​ (Daily Profit) | ​$5800 | $7320 | ​$4120

+y=ax%5E2%2Bbx%2Bc+........plug in x=30​ and y=5800​

+5800=a%2A30%5E2%2Bb%2A30%2Bc+
+5800=900a%2B30b%2Bc+..........solve for c
+c=5800-900a-30b+.......eq.1


+y=ax%5E2%2Bbx%2Bc+........plug in x=50​ and y=7320​

+7320=a%2A50%5E2%2Bb%2A50%2Bc+
+7320=2500a%2B50b%2Bc+..........solve for c
+c=7320-2500a-50b+.......eq.2

+y=ax%5E2%2Bbx%2Bc+........plug in x=100​ and y=4120​

+4120=a%2A100%5E2%2Bb%2A100%2Bc+
+4120=10000a%2B100b%2Bc+..........solve for c
+c=4120-10000a-100b+.......eq.3

from eq.1 and eq.2 we have
5800-900a-30b+=7320-2500a-50b......solve for+b
50b-30b+=7320-2500a-5800%2B900a
20b+=1520-1600a
b+=76-80a...........eq.1a

from eq.2 and eq.3 we have
7320-2500a-50b=4120-10000a-100b .....solve for b
100b-50b=4120-10000a-7320%2B2500a
50b=-7500a-3200
b=-150a-64..........eq2a

from eq.1a and eq.2a we have

76-80a=-150a-64.....solve for+a
150a-80a=-76-64
70a=-140
a=-140%2F70
a=-2
go to b+=76-80a...........eq.1a, substitute a
b+=76-80%28-2%29
b+=76%2B160
b+=236

go to +c=4120-10000a-100b+.......eq.3, substitute a and b
+c=4120-10000%28-2%29-100%2A236+
+c=4120%2B20000-23600+
+c=4120-3600+
+c=520+

so, your function is:+y=-2x%5E2%2B236x%2B520+
max is at vertex, so write your function in vertex form y=a%28x-h%29%5E2%2Bk where h and k are coordinates of the vertex


+y=-2x%5E2%2B236x%2B520+....using completing square we have
+y=-2%28x%5E2-118x%29%2B520+
+y=-2%28x%5E2-118x%2Bb%5E2%29-%28-2b%5E2%29%2B520+....b=118%2F2=59
+y=-2%28x%5E2-118x%2B59%5E2%29%2B2%2A59%5E2%2B520+
+y=-2%28x-59%29%5E2%2B6962%2B520+
+y=-2%28x-59%29%5E2%2B7482+
=>h=59 and k=7482

Each day 59 units should be produced to have a maximum daily profit of ​$7482.