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Jacob currently grows and sells 10 butternuts each week to a spaza shop. Each butternut costs R4 to grow and is sold for R7.

Jacob notices that the spaza shop does not yet sell cauliflowers. Cauliflowers cost R6 each to grow. Jacob visited the spaza shop and discussed the possibility of selling cauliflowers with the owner. He is open to the idea and is prepared to buy one cauliflower a week for R12,50. He could buy more cauliflowers, but is then prepared to pay less for each of the cauliflowers as he would not be able to sell them all at such a high price. After thinking about it, he is prepared to buy the following quantities of cauliflowers per week at the following prices:

Prepared to buy 1 2 3 4 5 6 7 8 9 10
Price per cauliflower bought R12.50 R12 R11.50 R11 R10.50 R10 R9.50 R9 R8.50 R8

Jacob can only grow and sell 10 vegetables in a week. What is the optimum number of cauliflower and butternut he should sell to maximize his profit?

Respuesta :

batolisis

The optimum number of cauliflower and butternut to maximize profit is : 6 Butternuts and 4 Cauliflowers

Using the Given data :

cost of producing 1 butternut = R4

selling price of 1 butternut = R7

profit from selling 1 butternut = R7 - R4 = R3

Assume: butternut sold in a week = x ,  Cauliflower sold in a week = y

Unit Cost of producing Cauliflower = R6

From the table on the amount of Cauliflower the shop owner is prepared to buy and at various unit prices

i) when the owner buys 4 Cauliflower

  unit profit = R11 - R6 = R5

ii) When shop owner buys 5 Cauliflower

   unit profit = R10.5 - R6 = R4.5

iii) when shop owner buys 10 Cauliflower

   unit profit = R8 - R6 = R2

Since Jacob can only grow and sell 10 vegetables in a week

The various combination of vegetables that he can grow and sell are as follows.

 x + y = 10       where ; profit on x = R3 , profit on y = R5,  R4.5, R2

i) 0 + 10 = 10     ------------- ( 1 )

  0 + 10(R2) =  R20

ii) 6 + 4 = 10    ---------------- ( 2 )

   6( R3 ) + 4(R5) = R38  ( maximum profit )

iii) 10 + 0 = 10  ------------ ( 3 )

   10(R3) + 0 =  R30

iv) 5 + 5 = 10   -------------- ( 4 )

  5( R3 ) + 5 ( R4.5 ) = R37.5

Hence from the combinations of Cauliflower and butternut the optimum number that would yield the highest profit for Jacob is ; 6 Butternut and 4 Cauliflower )

Learn more : https://brainly.com/question/22720939

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