Maximization and minimization
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 $100$ €.
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:
$x\geqslant 0$, $y\geqslant 0$ (the number of kilos cannot be negative).
$8x+4y\geqslant 16$ (at least we have to get $16$ g of the first substance).
$x+y\leqslant 5$ (at most we have to get $5$ g of the second substance).
$2\cdot x+2\cdot y\leqslant 20$ (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).
$(0,4)$ where cross the restrictions $x\geqslant 0$ and $8x+4y\geqslant 16$.
$(0,5)$ where cross the restrictions $x\geqslant 0$ and $x+y\leqslant 5$.
$(\dfrac{10}{3},\dfrac{5}{3})$ 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.
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 $112$ €.