During a​ war, allies sent food and medical kits to help survivors. Each food kit helped 8 people and each medicine kit helped 4 people. Each plane could carry no more than 50,000 pounds. Each food kit weighed 20 pounds and each medicine kit weighed 10 pounds. In addition to the weight constraint on its​cargo, each plane could carry a total volume of supplies that did not exceed 4000 cubic feet. Each food kit was 1 cubic foot and each medical kit also had a volume of 1 cubic foot. Assume that those helped by medicine kits were not helped by the food kits and vice versa. What was the maximum number of people that could be helped with one plane of​ supplies? The maximum number of people that could be helped was ___ people

Food kit and Medical kit are to be shipped to a survivor of a war by plane. Each container of Food kit will help 8 persons and each Medical kit will help 4 people. If 'x' represents the number of Food kit and ' y ' represents the number of Medical kit and ' z ' represents the number of people helped the objection function then
z=8x+6y
Each plane can be carried not more than 50000 pounds. The Food kit weighs 20 pounds and each Medical kit weighs 10 pounds. Let ' x ' be the number of Food kit and ' y ' be the number of Medical kit then
$20x+10y\le <!--\; \le \; -->50000$
Each plane carries a total volume of a supplier that does not exceed 4000 cubic feet. Each Food kit is 1 Cubic feet and each Medical kit is 1 cubic feet. Let ' x ' be representing the number of Food kit and ' y ' representing the number of Medical kit then
$1x+1y\le <!--\; \le \; -->4000$
In summary, here’s what we have described so far in this aid-to-war situation:
$$\begin{gathered} z=8x+6y \\ 20x+10y\le <!--\; \le \; -->50000 \\ 1x+1y\le <!--\; \le \; -->4000 \end{gathered}$$
We must maximize z=8x+6y subject to the following constraints:
$$\begin{gathered} 20x+10y\le <!--\; \le \; -->50000 \\ 1x+1y\le <!--\; \le \; -->4000 \end{gathered}$$
We use the addition method to find where the lines $20x+10y\le <!--\; \le \; -->50000and1x+1y\le <!--\; \le \; -->4000$ intersect.
20x+10y=50000
x+y=4000
If we solve the above 2 equations then we will get x=1000 and y=3000
Now we will substitute the values of x and y in z=8x+6y then
z= 8(1000)+6(3000)=8000+18000=36000
The maximum number of people that could be helped was 36000 people