Doubling Production. In general, the costs for doubling production of any type o

Question

Doubling Production. In general, the costs for doubling production of any type of item are expected to decrease b y approximately 25% per item. Ifwe know that the costs for producing 10,000 Bouncy Balls is $1500, and ifcosts decrease by25% foreach item when production is doubled, determine the cost equation for producing x thousand Bouncy Balls. n 1) What is the cost per item when 10,000 Bouncy Balls are prodhucedg when o,o00 Dalls are prod the totats soo Coct of produsing ench ball 0.IS Soo( IS 2) What is the cost per thousands of items when To,000 Bouncy Balls are bails are produci the total cost is 4ISoo produced? $ISo when to,000 balls are produced the total cost is cost of Producing each fnuandskalls ISoo/lo: $160 3) Fill in the table of Values below and use the results to find the cost equation. r (thousand) Bouncy Balls Calculation for Cost of Production Cost # items . Cost per Item 10(1500 1500 10 = 20 . 10 20 = 21.10 40=22.10 80 = 23 . 10 3219 x=2?? . 10 10(ISO , In order to fill in the last row of the table, you will need to solve the following equation for y: 4) x = 2, 10. Give the solution to the equation. n=

Explanation / Answer

Your calculation so far is correct:
Ans(4)
Look at the table
we see that first column increases according to formula (2^n)*10
2nd column is product of 1st column and (150* (3/4)^n)
third column is just simplification
so the formula for cost that is formula that goes into 2nd column of the last row will be:
[(2^n)*10]*[150* (3/4)^n]
Hence required equation is
cost = y = [(2^n)*10]*[150* (3/4)^n]
where x=(2^n)*10
or
x/10=2^n
or log(x/10)=nlog(2)
or n=log(x/10)/log(2)
Hence final answer for question 4 is
n=log(x/10)/log(2)

------------
As per Chegg answering rule you need to post other parts separately to get then answer.
Thanks

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.