Question

There are two truck kinds available to a transportation company: Type A and Type B. When comparing Type A with Type B, Type A has an overall volume of 20 m3 with equal portions for refrigerated and non-refrigerated stock, while Type B has an overall volume of 40 m3 with equal sections for both types of stock. 3,000 m3 of refrigerated stock and 4 000 m3 of nonrefrigerated goods must be transported by truck. Type A cars cost P30 per kilometer, whereas Type B cars cost P40 per kilometer. In order to keep rental costs to a low, how many of each kind of truck should the grocer rent?

Answer 1

Lets truck type A - $x$ , truck type B - $y$ .

Type A has an overall volume of 20 m³ with equal portions for refrigerated (10 m³ ) and non-refrigerated stock (10 m³).

Type B has an overall volume of 40 m³ with equal for refrigerated (20 m³ ) and non-refrigerated stock (20 m³).

3000 m³ of refrigerated stock

$10x+20y\geq3000$

4 000 m³ of nonrefrigerated stock

$10x+20y\geq4000$

Type A cars cost P30 per kilometer, Type B cars cost P40 per kilometer.

$F(x,y)=30x+40y\to min$

$10x+20y\geq3000\\ 10x+20y\geq4000\\ F(x,y)=30x+40y\to min\\ x+2y\geq300\\ x+2y\geq400\\ x\geq0,y\geq0, x\in N, y\in N\\ x+2y\geq400\\ x+2y=400\\ \begin{matrix} x& 0 &200\\ y& 20&100 \end{matrix}\\ 30x+40y=0\\ \begin{matrix} x& 0 &4\\ y& 0&-3 \end{matrix}\\ \vec{a}=(30,40)$

In parallel, transfer the line $30x+40y=0$ in the direction of the vector $\vec{a}=(30,40)$ . We are looking for a point of intersection with a straight line $x+20y=400$

We have a point (109.91, 145,455), $x\in N, y\in N\implies$ $x=110, y=146$