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maxonik [38]
2 years ago
5

A manufacturer sells each of his TV sets for $85. The cost C (in dollars) of manufacturing and selling x TV sets per week is C =

1500 + 10x + 0.005x2. If at most 10,000 sets can be produced per week, how many sets should be made and sold to maximize the weekly profit?
Mathematics
2 answers:
igor_vitrenko [27]2 years ago
7 0
Answer: You should produce all 10,000 TV sets to reach the maximum profit.

If you graph the equation for the profit (below), you will see that the function is increasing on the entire interval from 0 to 10,000.
P = 85x - (1500 + 10x + 0.005x^2)

Therefore, the maximum project would be at the end of the interval, or 10,000 TV units.
romanna [79]2 years ago
7 0

Answer:

7500 sets

Step-by-step explanation:

We are given that

Cost of each TV set=$85

The cost C(in dollars) of manufacturing  and selling x Tv sets per week is given by

C=1500+10x+0.005x^2

Set can be produced at most per week=10000

We have to find the number of sets should be made and sold to maximize the weekly profit.

Revenue=85 x

Where x= number of sets selling per set

Profit=Revenue-Cost

Profit=85x-(1500+10x+0.005x^2)[/tex]

Profit=P=85x-1500-10x-0.005x^2

Differentiate w.r.t x

\frac{dP}{dx}=85-10-0.010x=75-0.010x

Substitute \frac{dP}{dx}=0

75-0.010x=0

0.010x=75

x=\frac{75}{0.010}=7500

Again differentiate w.r.t x

\frac{d^2P}{dx^2}=-0.010 < 0

Hence,function is maximize.

Therefore, 7500 sets should be made and sold to maximize the weekly profit.

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Answer:

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Step-by-step explanation:

If the total games played was 36 and no team played each other twice, we need to ensure there isn't any double counting.

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using this knowledge, we can then count up:

1+2+3+4+5+6+7+8 = 36

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2 years ago
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You can do this if you convert all the measures into one unit. Let us convert all into km. 

The first one is already in km, so we do not need to covert it. 

Let's start with converting 1,220,000mm to km. 


There are 1, 000,000 mm in one km. 

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