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Two practical applications of coordinate geometry (higher level courseware)
How can we gain the highest profit?
In economics it is very important to plan and organise certain production processes in the most economical way possible. The following example shows such a method used very often for optimal planning.
Example 1
One of the plants of a multinational company produces CD-ROM drives forpersonal computers. The plant produces two types of CD-ROM drives, let these be \latex{D_{1}} and \latex{D_{2}.}The assembly of a product consists of three consecutive phases. The assembly operations corresponding to the phases are done by three distinct automatic production lines (let these be denoted by \latex{ A }, Band \latex{ C } according to the order of production) so that all three production lines are suitable for the production of both types of device but under different conditions:
- The production line Aprepares \latex{ 1 } piece of \latex{D_{1}} in \latex{ 5 } minutes, while it prepares \latex{ 1 } piece of \latex{D_{2}} in \latex{ 10 } minutes for the next production phase, and it canbe operated for at most \latex{ 16 } hours daily.
- The production line \latex{ B } prepares \latex{ 1 } piece of \latex{D_{1}} and \latex{D_{2}} both in \latex{ 5 } minutes for the last phase; however it can be operated for at most \latex{ 9 } hours daily.
- The production line \latex{ C } can be operated non-stop, i.e. for \latex{ 24 } hours daily, and it produces \latex{ 1 } piece of \latex{D_{1}} in \latex{ 20 } minutes and \latex{ 1 } piece of \latex{D_{2}} in \latex{ 5 } minutes.
The product \latex{D_{1}} produces a net profit of €\latex{ 3 }, and the product \latex{D_{2}} produces a net profit of €\latex{ 2 } per piece for the plant when discounting expenses (tax, overhead expenses, wages, etc.).
How shall the daily production be organised in the plant if the aim is to gain the highest profit possible?
How shall the daily production be organised in the plant if the aim is to gain the highest profit possible?
Solution
For the better understanding let us tabulate the data. (For the sake ofsimplicity let the unit of time be \latex{ 5 } minutes.)
production
line
line
production time
of \latex{ 1 } piece of \latex{D_{1}}
of \latex{ 1 } piece of \latex{D_{1}}
production time
of \latex{ 1 } piece of \latex{D_{2}}
of \latex{ 1 } piece of \latex{D_{2}}
maximum operation time
of the production line
of the production line
\latex{ A }
\latex{ B }
\latex{ C }
\latex{ 1 }
\latex{ 2 }
\latex{ 192 }
\latex{ 1 }
\latex{ 1 }
\latex{ 108 }
\latex{ 4 }
\latex{ 1 }
\latex{ 288 }
Let x denote the number of \latex{D_{1}} products produced in \latex{ 1 } day, let \latex{ y } denote the number of \latex{D_{2}} products produced in \latex{ 1 } day (\latex{ x } and \latex{ y } are obviously whole numbers). Based on the conditions the following inequalities need to hold for \latex{ x } and \latex{ y }:
- \latex{x\geq 0,}
- \latex{y\geq 0,}
- \latex{x+2y\leq 192,}
- \latex{x+y\leq 108,}
- \latex{4x+y\leq 288.}
We are looking for the solutions \latex{ x } and \latex{ y } of this system of inequalities for which the expression \latex{h=3x+2y} takes the largest value, i.e. the value of the daily profit is the highest.
Each of the inequalities (1) to (5) defines a half-plane in the Cartesian coordinate system. The intersection of these half-planes will be a range the coordinates of the points \latex{P(x;y)} of which and only these satisfy all of the inequalities (1) to (5). Let us represent this range.
The first quadrant of the coordinate system corresponds to the combi nation of the inequalities (1) and (2) (Figure 81).
The first quadrant of the coordinate system corresponds to the combi nation of the inequalities (1) and (2) (Figure 81).
\latex{ y }
\latex{ 10 }
\latex{ 0 }
\latex{ 10 }
\latex{ x }
Figure 81
\latex{x+2y=192}
\latex{ 96 }
\latex{ 10 }
\latex{ 0 }
\latex{ 10 }
\latex{ 192 }
\latex{ x }
\latex{ y }
Figure 82
The inequality (3) defines the range “below” the straight line with the equation \latex{x+2y=192.} The range corresponding to the inequalities (1), (2) and (3) can be seen in Figure 82.
The range corresponding to the inequalities (1), (2), (3) and (4) can be seen in Figure 83, and the range corresponding to all five inequalities can be seen in Figure 84.
The range corresponding to the inequalities (1), (2), (3) and (4) can be seen in Figure 83, and the range corresponding to all five inequalities can be seen in Figure 84.
\latex{x+2y=192}
\latex{x+y=108}
\latex{ 96 }
\latex{ 108 }
\latex{ y }
\latex{ 10 }
\latex{ 0 }
\latex{ 10 }
\latex{ 108 }
\latex{ 192 }
\latex{ x }
Figure 83
\latex{4x+y=288}
\latex{x+2y=192}
\latex{x+y=108}
\latex{ 96 }
\latex{ 108 }
\latex{ y }
\latex{ 10 }
\latex{ 0 }
\latex{ 10 }
\latex{ 72 }
\latex{ 108 }
\latex{ 192 }
\latex{ x }
Figure 84
The inequalities (1) to (5) hold at once for the coordinates of the points with integer coordinates belonging to the closed convex pentagon shown in Figure 84. We have to determine for which of these points the expression \latex{h=3x+2y} is maximal.
In Figure 85 it can be seen that the expression above describes a set of parallel straight lines, and it can also be seen that \latex{ h } is maximal for the straight line that passes through the intersection point of the straight lines with the equation \latex{x+y=108} and \latex{4x+y=288.} By solving the simultaneous equations consisting of the equations of the two straight lines we find that it is the point \latex{P(60;48).}
\latex{P(60;48)}
\latex{3x+2y=0}
\latex{3x+2y=80}
\latex{3x+2y=160}
\latex{3x+2y=276}
\latex{ 96 }
\latex{ y }
\latex{ 0 }
\latex{ 10 }
\latex{ 72 }
\latex{ x }
Figure 85
It means that the plant gains the highest profit if it produces \latex{ 60 } pieces of the product \latex{D_{1}} and \latex{ 48 } pieces of the product \latex{D_{2}} daily. Then the daily profit is
\latex{h=3\times 60+2\times 48=€276.}
It can also be calculated that the machines \latex{ B } and \latex{ C } work for the maximum allowed time, while the machine \latex{ A } works \latex{ 14 } hours daily to gain the maximal profit.
The problem just discussed belongs to the topic of linear programming, and the solution method showed here is called the graphical method.
The problem just discussed belongs to the topic of linear programming, and the solution method showed here is called the graphical method.
How does a dish aerial work and why does it work that way?
Dish aerials are very important tools of today's telecommunication. They receive and transmit electromagnetic signals radiated by the satellites orbiting synchronised with the rotation of the Earth to the corresponding communication devices (e.g. television, mobile phone, computer).
If we rotate a parabola about its axis of symmetry, then a paraboloid of revolution is obtained. The dish of a dish aerial is the surface obtained when intersecting a paraboloid of revolution with a plane perpendicular to the axis of revolution (Figure 86).
The idea behind the operation of a dish aerial is that the signals arriving at its dish parallel with its axis of revolution are reflected back from the surface, and the reflected signals get to the receiver unit (detector) placed at the focus of the paraboloid (Figure 87). This phenomenon can be explained with the law of reflection and a beautiful geometric property of the parabola. Because of the rotational symmetry it is enough to examine a plane section coinciding the axis (Figure 87).
Figure 86
Let the signal parallel with the axis (the straight line \latex{ b } in Figure 88) arrive at the point \latex{ E } of the parabola. It is then reflected (straight line \latex{ v }) so that the angle of incidence and the angle of reflection included between the rays and the straight line perpendicular to the tangent line e passing through \latex{ E } of the parabola are equal. But then the angles included between \latex{ b } and \latex{ e }, and \latex{ v } and \latex{ e } are also equal, i.e. the tangent line bisects one of the angles included between the straight lines \latex{ b } and \latex{ v }. This fact is a consequence of the law of reflection.
Now we are going to show using coordinate geometry in a particular case that the parabola has a property that assures that the straight line v passes through the focus \latex{ F }.
Now we are going to show using coordinate geometry in a particular case that the parabola has a property that assures that the straight line v passes through the focus \latex{ F }.
\latex{ F }
\latex{ t }
Figure 87
THEOREM: The tangent line drawn to an arbitrary point \latex{ P } of the parabola bisects the angle included between the straight line connecting the focus and \latex{ P } and the straight line perpendicular to the directrix and passing through \latex{ P }.
Proof (for a particular parabola)
Let us consider the parabola with the equation \latex{y=x^{2}}, and let \latex{ e } be the tangent line drawn to the point \latex{P(a;a^{2} )(a\gt 0)} (Figure 89).
The focus of the parabola is \latex{F (0;\frac{1}{4} )}, the equation of its directrix is \latex{y=-\frac{1}{4}.}
The straight line with the equation \latex{x = a} intersects the directrix at the point \latex{T (a;-\frac{1}{4} ).}
The focus of the parabola is \latex{F (0;\frac{1}{4} )}, the equation of its directrix is \latex{y=-\frac{1}{4}.}
The straight line with the equation \latex{x = a} intersects the directrix at the point \latex{T (a;-\frac{1}{4} ).}
\latex{ E }
\latex{ b }
\latex{ t }
\latex{ v }
\latex{ e }
Figure 88
\latex{x=a}
\latex{P(a;a^{2} )}
\latex{F(0;\frac{1}{4} )}
\latex{y=-\frac{1}{4}}
\latex{T(a;-\frac{1}{4} )}
\latex{ e }
\latex{ 1 }
\latex{ 0 }
\latex{ 1 }
x
f
Figure 89
The definition of the parabola implies that \latex{ FP = PT }, i.e. the triangle \latex{ FTP } is isosceles. We are going to prove that \latex{ e } is perpendicular to the straight line \latex{ FT }, which also means that it bisects the angle \latex{ TPF }.
The steps of the proof are:
The steps of the proof are:
- Determining the gradient of \latex{ e }.
- Determining the gradient of the straight line\latex{ FT }.
- Comparing the two gradients.
- Let the gradient of e be denoted by \latex{ m }. Thus the equation of \latex{ e } is:
\latex{y-a^{2}=m\times (x-a).}
\latex{ e } has exactly one point in common with the parabola, therefore the simultaneous equations
\latex{\begin{rcases}y=x^{2} \\ y-a^{2}=m\times (x-a)\end{rcases}}
have one solution. It is satisfied exactly when the discriminant of the quadratic equation
\latex{x^{2}-mx+ma-a^{2}=0}
is \latex{ 0 }, i.e.
\latex{m^{2}-4ma+4a^{2}=(m-2a )^{2}=0.}
This implies that
\latex{m=2a}
- The gradient m’ of the straight line \latex{ FT } is:
\latex{m'= \frac{-\frac{1}{4}-(-\frac{1}{4} ) }{0-a}=\frac{1}{2a}.}
- As \latex{m\times m'=-1,} e intersects \latex{ FT } indeed perpendicularly, thus it bisects the angle \latex{ TPF }.
Thus we have shown the theorem for this particular case.
Note: In Figure 89 it can be seen but with calculation it can also easily be verified that the tangent line \latex{ e } and the straight line \latex{ FT } intersect each other at the point \latex{M (\frac{a}{2};0 ).}
Note: In Figure 89 it can be seen but with calculation it can also easily be verified that the tangent line \latex{ e } and the straight line \latex{ FT } intersect each other at the point \latex{M (\frac{a}{2};0 ).}
