Question 318404
Let X be the number of hours using Machine A.
Let Y be the number of hours using Machine B.
Total quantity:{{{30X+40Y=380}}}
{{{3X+4Y=380}}}
Total costs:{{{8X+12Y<=108}}}
You must satisfy both equations : making the number of steering wheels and keeping cost below or equal to a certain limit.
Find the point of intersection between the two curves, 
1.{{{3X+4Y=38}}} 
2.{{{8X+12Y=108}}}
Multiply eq. 2 by (-3) and add to eq. 1,
{{{-9X-12Y+8X-12Y=-114+108}}}
{{{-X=-6}}}
{{{highlight( X=6 )}}}
.
.
.
Then use either equation to solve for Y,
{{{3(6)+4Y=38}}}
{{{4Y=20}}}
{{{highlight(Y=5)}}}
.
.
.
Run machine A for 6 hours, machine B for 5 hours to make 380 steering wheels at a total cost of $108. 
.
.
.
Additionally, this is actually a linear programming problem. 
We should better define the feasible region, although the region is not a region but a line segment.
The red line shows the cost limit while the green line shows the number of steering wheels. 
Once the green line crosses the red line, the total cost inequality does not hold. 
So the only points we need to check are the intersection point between the red and green curve and the intersection point between the green line and the x-axis.
What we're trying to minimize is the number of hours total, {{{X+Y}}}
At {{{X=6}}}, {{{Y=5}}}, {{{X+Y=11}}}
When {{{Y=0}}}, {{{3X=38}}}, {{{X=38/3}}} and then {{{X+Y=38/3=12.7}}}
So the minimum for {{{X+Y}}} does occur at {{{X=6}}}, {{{Y=5}}}
{{{drawing(300,300,-2,16,-2,16,circle(6,5,0.3),circle(12.67,0,.3),grid(1),graph(300,300,-2,16,-2,16, (108-8x)/12,(38-3x)/4))}}}

*[invoke plot_any_inequality "y < (108-8x)/12", -2,16, -2, 16, 300, 300]