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Mark has opened a store and is selling wagons. He has a capital of 416 000 dollars that he can use to purchase 2 different type of wagons. Model A has a purchase price of 2400 dollars and a profit at sale of 1000 dollars. Model B has a purchase price of 4000 dollars and a profit at sale of 1200 dollars. The storage in the store allows that the purchase consist of a maximum of 150 wagons. What is the largest profit you can do with the disposable purchase amount?

This is the question, can someone help me or tell me what function I should use to solve it? any help is appreciated!!!!!

Thank you for your time guys for helping me out. This is what I did so far. Can you tell me if I did it correct?

x + y = 150 | * (-4000)

2400x + 4000y = 416 000



-4000x - 4000y = -600 000

2400x + 4000y = 416 000



-1600x = -184 000

x = -184 000 / -1600

x = 115



115 + y = 150 

y = 150 - 115

y = 35



1000 * 115 = 115 000

1200 * 35 = 42 000



115 000 + 42 000 = 157 000 dollars
callculus42
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J. Doe
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  • Hint: Lagrange multipliers. Let $x$ be the number of model A purchased and $y$ be the number of model B purchased, and then write a relationship between $x,y$ considering how much money Mark can use. Afterwards, write a profitability function and find the maximum of that function on the relationship space. – AnonymouseCat May 20 '16 at 12:40
  • Are they looking for a function like this?

    x + y = 150 2400x + 4000y = 416000 1000x + 1200y = z

    first demonstrates that both wagons has to have a maximum of 150 wagons together second demonstrates that purchase price * amount of wagons = total investment money third demonstrates profit per item * amount = total profit

    like all 3 are connected and you count them into 1 equation?

     – J. Doe May 20 '16 at 12:56

1 Answers1

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Mark's order will be $(a,b)$, meaning $a\geq0$ wagons of type A and $b\geq0$ wagons of type B. Draw a figure showing the first quadrant of the $(a,b)$-plane. There are two conditions, namely $$a+b\leq 150, \qquad 2.4\,a+4b\leq 416\ .$$ These conditions define an admissible region $A$, a quadrilateral in the $(a,b)$-plane bounded by the axes and two lines $g$, $h$ with slope $-1$ and $-0.6$ respectively.

On the other hand, his profit $p(a,b)$ (in thousands of dollars), if he can sell all his wagons, will be $$p(a,b)=a+1.2 b\ .$$ All points on a line $\ell: \>a+1.2\, b={\rm const.}$ of slope $-{5\over6}$ will lead to the same profit, and the further north-east this line is, the higher is the profit. In order to find the point on $A$ leading to the highest profit we therefore have to translate such a line from far north-east in direction south-west, until it for the first time hits a point of $A$. As $0.6<{5\over6}<1$ this point will be the point of intersection $g\wedge h$. Compute this point, and you will know what Mark has to do.

Christian Blatter
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