Answer
The maximum number of 300 cartons of food and 200 cartons of clothes per shipment can be sent at a time.
Work Step by Step
Given that, one carton of food feeds 12 people and one carton of clothing helps 5 people.
Say x number of cartons of food and y number of cartons of clothing per plane shipment are enough to provide maximum help to the people.
Possible objective function to define help provided to maximum number of people can be represented as
\[z=12x+5y\]
The weight constraint about the number of food boxes and clothing boxes results in
\[\begin{align}
& 50x+20y\le 19,000 \\
& 5x+2y\le 1900
\end{align}\]
The volume constraint about the number of boxes results in
\[\begin{align}
& 20x+10y\le 8000 \\
& 2x+y\le 800
\end{align}\]
The number of boxes for food and clothing is a non-negative number and provides another constraint as
\[\begin{align}
& x\ge 0 \\
& y\ge 0
\end{align}\]
Now,try to maximize the value of objective function\[z=12x+5y\] according to the above constraints.
For this, try to find out the vertex points from the linear equations of the above inequalities viz. \[2x+y=800,\text{ }5x+2y=1900,\text{ }x=0,\text{ }y=0\].
Set 0 for y in the equation\[2x+y=800\]to get
\[\begin{align}
& 2x+0=800 \\
& 2x=800 \\
& x=400
\end{align}\]
This provides x-intercepts as 400 and we get the vertex point as \[\left( 400,0 \right)\].
Set \[0\]for y in the equation\[5x+2y=1900\]to get
\[\begin{align}
& 5x+2.0=1900 \\
& 5x=1900 \\
& x=380
\end{align}\]
Thus,we get x-intercepts as 380 and we get the next vertex point as \[\left( 380,0 \right)\].
Now, set 0 for x in the equation\[2x+y=800\]to get
\[\begin{align}
& 2.0+y=800 \\
& 0+y=800 \\
& y=800
\end{align}\]
We get the y-intercepts as 800 and we get the next vertex point as \[\left( 0,800 \right)\].
Setting 0 for x in the equation\[5x+2y=1900\]gives
\[\begin{align}
& 5.0+2y=1900 \\
& 0+2y=1900 \\
& y=950
\end{align}\]
So,we get the next y-intercepts as 950and another vertex point as \[\left( 0,950 \right)\].
Vertex point \[\left( 0,0 \right)\]is the result of the intersection point of axis lines\[x=0,y=0\].
One more vertex point is possible from the intersection points of equations\[2x+y=800\text{ and }5x+2y=1900\].
Solve these equations together by multiplying the first equation by \[-2\] and then adding it to the second equation as follows:
\[\begin{align}
& -2\left( 2x+y \right)+5x+2y=-2\left( 800 \right)+1900 \\
& -4x-2y+5x+2y=-1600+1900 \\
& x=300
\end{align}\]
Back substitute\[x=300\]in\[2x+y=800\] to obtain
\[\begin{align}
& 2\left( 300 \right)+y=800 \\
& 600+y=800 \\
& y=200
\end{align}\]
Another vertex point is\[\left( 300,200 \right)\].
Consider a test point\[\left( 0,0 \right)\]. This will be used to determine the shaded region.
For \[2x+y\le 800\], we have
\[\begin{align}
& 2\left( 0 \right)+\left( 0 \right)\le 800 \\
& 0\le 800
\end{align}\]
As this inequality holds true, the shaded region would be toward the origin.
For \[5x+2y\le 1900\], we have
\[\begin{align}
& 5\left( 0 \right)+2\left( 0 \right)\le 1900 \\
& 0\le 1900
\end{align}\]
As this inequality holds true, the shaded region would be toward the origin.
