I have a question as shown above (task4 part iii) which requires us to determine

Question

I have a question as shown above (task4 part iii) which requires us to determine the output level where marginal cost are mimimized.

to find the output level, i have divided the solution into the following steps:

Step 1: determine the MC function

MC = dTC/dQ
      = d(0.10Q3 - 3Q2 + 60Q + 250)/dQ
      = 0.3Q2 -6Q +60

Step 2: for MC to minimize, it means MC =dTC/dQ =0

therefore ,

0.3Q2-6Q+60 =0

my enquiry is this, i'm not sure how to proceed from here.. is this approach correct? what is actually the value for Q since i foresee there will be two Q values.

appreciate your help with rgrds to this tq

CHAPTER AND OLIG CHAPTER 7: COST ANALYSIS TASK4: Noks oco of production TASK 5 PURPOSE The purpose of Task 4 is to enhance students' knowledge and and the optimisation of the use of variable input. PURPOSE The purp oligopoly the application of the cost an REQUIREMENT REQUIRE In order to maintain its competitiveness among other oyste market, the production manager of Maggi Oyster Sauce cost effectiveness of the company's production. The to relationship TC 0.10Q3-3Q2 +60Q+250 1. r s 1. (a) ng sfhee of Maggi Oystetrh er oyster sauc producers in the Ma The total costs are given by the follow where TC= total cost Qquantity of oyster sauce produced (in 000) Derive an average total cost function. Determine the output level where average variable costs are minimized ii. Determine the output level where margin iv. What is the marginal cost when 10,000 bottles of oyster sauce are produced? Does al costs are minimized 2. it the same if the number of production increase to 11,000 bottles? As the production manager, what should you do? Give two reasons why is marginal cost (MC) important for a firm? v.

Explanation / Answer

In your solution, you have taken the first order condition of Tc, and not MC. That is why you are unable to proceed further from the equation. the correct solution is as follows:

Given TC = 0.10Q^3 - 3Q^2 + 60Q + 250

MC = dTC/dQ = 0.30Q^2 - 6Q + 60

and ATC = TC/Q = 0.10Q^2 - 3Q + 60 + 250/Q

(iii) In order to minimize MC, we take the first order condition: dMC/dQ = 0

i.e. 0.60Q - 6 = 0 which gives Q = 10.

Taking the second order condition of minimization, d^2 MC/dQ^2 = 0.60 > 0, thus MC is minimized.

Ans. Q = 10 and MC(at Q=10) = 30.

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