We have the cost function in function of the number of hamburgers as:C
![C(h)=4000+8h](https://img.qammunity.org/2023/formulas/mathematics/college/o4oh2zma6f2o23fxan9v947dxoa3y18ol2.png)
1.1) We have to represent the income I.
It will be equal to the number of hamburgers times the price of each hamburguer ($20).
Then, we can write it as:
![I(h)=20h](https://img.qammunity.org/2023/formulas/mathematics/college/6xg62sl89682kf6hyualx0ozck2m5d2220.png)
1.2) We have to complete the table for each production level.
We can use the functions to calculate for each level.
We start with the expenses:
![\begin{gathered} C(0)=4000+8(0)=4000+0=4000 \\ C(200)=4000+8(200)=4000+1600=5600 \\ C(400)=4000+8(400)=4000+3200=7200 \\ C(600)=4000+8(600)=4000+4800=8800 \\ C(800)=4000+8(800)=4000+6400=10400 \\ C(1000)=4000+8(1000)=4000+8000=12000 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/fou6uzr1u16cv6ggo8jmm7zc5sr6qm9hx0.png)
We can now calculate the income I(h) as:
![\begin{gathered} I(0)=20(0)=0 \\ I(200)=20(200)=4000 \\ I(400)=20(400)=8000 \\ I(600)=20(600)=12000 \\ I(800)=20(800)=16000 \\ I(1000)=20(1000)=20000 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/xukz5vclkk1yqd1tp3z3fgyy2ofe8fpgfz.png)
We can then complete the table as:
We now have to graph both functions.
We can use the horizontal axis for h and the vertical axis for C and I and obtain:
1.4) From the graph we know that the amount of hamburguers he need to sell to break even is between 300 and 400, where C(h) = I(h).
We can use the equations to find the exact value:
![\begin{gathered} C(h)=I(h) \\ 4000+8h=20h \\ 4000=(20-8)h \\ 4000=12h \\ h=(4000)/(12) \\ h\approx334 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/zzqrbqw21a58n4asfyufj1luw2pw9mggsf.png)
1.5) We can calculate how much he loss when h = 200 as the difference between the income and the cost for that level.
![\begin{gathered} P(200)=I(200)-C(200) \\ P(200)=4000-5600 \\ P(200)=-1600 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/y8mnr3fmcgoldt7t9965mqxcyptydu8026.png)
He loss $1600 if he sells only 200 hamburguers.
1.6) We have to calculate h for I(h) = 15000:
![\begin{gathered} I(h)=15000 \\ 20h=15000 \\ h=(15000)/(20) \\ h=1250 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ezh2rhjn3o3sy6ib0s3l04hkfgoey11z3l.png)
The number of hamburguers is 1250.
1.7) The profit for 1000 hamburguers can be calculated as:
![\begin{gathered} I(1000)-C(1000) \\ 20000-12000 \\ 8000 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/1lmj89fhn412hjoteys0a6x1z5atxlarzl.png)
The maximum profit would be $8000.
Answer:
1.1) I(h) = 20h
1.2) Given in the answer
1.4) 334 hamburguers
1.5) He will loss R 1600
1.6) 1250 hamburguers
1.7) R 8000