Question 549747
{{{x}}} cars take up {{{20x}}} of space (in {{{m^2}}})
{{{y}}} trucks take up {{{100y}}} of space (in {{{m^2}}})
If {{{x}}} cars and {{{y}}} trucks will fit into {{{1200}}}{{{m^2}}}
{{{20x+100y<=1200}}} <--> {{{x+5y<=60}}} dividing everything by 20.
At least 40 vehicles translates into {{{x+y>=40}}}
At least 2 trucks translates into {{{y>=2}}}
The inequalities asked for are
{{{x+5y<=60}}}, {{{y>=2}}} and {{{y>=2}}}
EXTRA THOUGHTS
It's only logical to ask that {{{x>=0}}} too, but that will happen anyway, as a consequence of the other restrictions above.
The inequalities determine a feasibility region. The boundaries for that feasibility region are:
{{{x+5y=60}}}, {{{x+y=40}}}, and {{{y=2}}}.
Those boundaries can be graphed as:
{{{graph(300,300,-10,90,-5,45,2,40-x,12-x/5)}}} The feasibility region is the tiny triangle between the 3 lines, sides included.