Problems from 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 $100$ €.

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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:

$x ⩾ 0$ , $y ⩾ 0$ (the number of kilos cannot be negative).
$8 x + 4 y ⩾ 16$ (at least we have to get $16$ g of the first substance).
$x + y ⩽ 5$ (at most we have to get $5$ g of the second substance).
$2 \cdot x + 2 \cdot y ⩽ 20$ (at most we have to get $20$ g of the third substance).
$x ⩽ 2 \cdot y$ (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 $2 \cdot x + 2 \cdot y ⩽ 20$ does not contribute excellent information, that is to say, it does not delimit the validity region).

$(0, 4)$ where cross the restrictions $x ⩾ 0$ and $8 x + 4 y ⩾ 16$ .
$(0, 5)$ where cross the restrictions $x ⩾ 0$ and $x + y ⩽ 5$ .
$(\frac{10}{3}, \frac{5}{3})$ where cross the restrictions $x + y ⩽ 5$ and $x ⩽ 2 \cdot y$ .
$(\frac{8}{5}, \frac{4}{5})$ where cross the restrictions $8 x + 4 y ⩾ 16$ and $x ⩽ 2 \cdot y$ .

Objective value of the function in the apexes of the area of validity:

$C (0, 4) = 20 \cdot 0 + 100 \cdot 4 = 400$
$C (0, 5) = 20 \cdot 0 + 100 \cdot 5 = 500$
$C (\frac{10}{3}, \frac{5}{3}) = 20 \cdot \frac{10}{3} + 100 \cdot \frac{5}{3} = \frac{700}{3} \approx 233.33$
$C (\frac{8}{5}, \frac{4}{5}) = 20 \cdot \frac{8}{5} + 100 \cdot \frac{4}{5} = \frac{560}{5} = 112$

The function cost has a minimal value ( $112$ €) at point $(\frac{8}{5}, \frac{4}{5})$ , that is to say, when we buy $\frac{8}{5}$ of kilo of the substance A and $\frac{4}{5}$ of kilo of the B.

Solution:

To manage to minimize the cost, taking into account the restrictions of the problem, $\frac{8}{5}$ of kilo of the substance A and $\frac{4}{5}$ of kilo of the B ought to be bought. In this case the cost would be $112$ €.

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