We want to obtain three chemical elements from the substances A and B. One kilo of A contains $$8$$ grams of the first element, $$1$$ gram of the second and $$2$$ of the third ; one kilo of B has $$4$$ grams of the first element, $$1$$ gram of the second one and $$2$$ of the third one. If we want to obtain at least $$16$$ grams of the first element and the quantities of the second one and of the third one have to be maximum $$5$$ and $$20$$ grams respectively, and the maximum quantity of A is double that of B, calculate the kilos of A and those of B that have to used so that the cost is minimal if one kilo of A costs $$20$$ € and one of B $$100$$ €.
Development:
Variables of the problem:
$$x$$: Kilos of the substance A.
$$y$$: Kilos of the substance B.
Objective Function:
The cost has to be minimized (cost $$=$$ (price of the kilo of the substance A) $$\times$$ (price of the kilo of A) $$+$$ (price of the kilo of the substance B) $$\times$$ (price of the kilo of B)): $$$C(x,y)=20\cdot x+100\cdot y$$$
Restrictions:
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$$x\geqslant 0$$, $$y\geqslant 0$$ (the number of kilos cannot be negative).
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$$8x+4y\geqslant 16$$ (at least we have to get $$16$$ g of the first substance).
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$$x+y\leqslant 5$$ (at most we have to get $$5$$ g of the second substance).
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$$2\cdot x+2\cdot y\leqslant 20$$ (at most we have to get $$20$$ g of the third substance).
- $$x\leqslant 2\cdot y$$ (the amount of substance A is at most twice the amount of B).
Apexes of the region of validity: (These are the intersection points between the straight lines associated with the restrictions, which also satisfy all the inequations. Note that the restriction $$2\cdot x+2\cdot y\leqslant 20$$ does not contribute excellent information, that is to say, it does not delimit the validity region).
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$$(0,4)$$ where cross the restrictions $$x\geqslant 0$$ and $$8x+4y\geqslant 16$$.
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$$(0,5)$$ where cross the restrictions $$x\geqslant 0$$ and $$x+y\leqslant 5$$.
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$$(\dfrac{10}{3},\dfrac{5}{3})$$ where cross the restrictions $$x+y\leqslant 5$$ and $$x\leqslant 2\cdot y$$.
- $$(\dfrac{8}{5},\dfrac{4}{5})$$ where cross the restrictions $$8x+4y\geqslant 16$$ and $$x\leqslant 2\cdot y$$.
Objective value of the function in the apexes of the area of validity:
- $$C(0,4)=20\cdot 0+100\cdot 4=400$$
- $$C(0,5)=20\cdot 0+100\cdot 5=500$$
- $$C(\dfrac{10}{3},\dfrac{5}{3})=20\cdot \dfrac{10}{3}+100\cdot \dfrac{5}{3}=\dfrac{700}{3}\approx 233.33$$
- $$C(\dfrac{8}{5},\dfrac{4}{5})=20\cdot \dfrac{8}{5}+100\cdot \dfrac{4}{5}=\dfrac{560}{5}=112$$
The function cost has a minimal value ($$112$$ €) at point $$ (\dfrac{8}{5},\dfrac{4}{5}) $$, that is to say, when we buy $$\dfrac{8}{5}$$ of kilo of the substance A and $$\dfrac{4}{5}$$ of kilo of the B.
Solution:
To manage to minimize the cost, taking into account the restrictions of the problem, $$\dfrac{8}{5}$$ of kilo of the substance A and $$\dfrac{4}{5}$$ of kilo of the B ought to be bought. In this case the cost would be $$112$$ €.