You are about to erase your work on this activity. Are you sure you want to do this?
Updated Version Available
There is an updated version of this activity. If you update to the most recent version of this activity, then your current progress on this activity will be erased. Regardless, your record of completion will remain. How would you like to proceed?
Mathematical Expression Editor
Introduction
Suppose a rectangle has width and length , with area and perimeter . The area of
the rectangle is and the perimeter is given by , giving the following system of
equations
Since the first equation here is not a linear equation, this is a nonlinear system of
equations. If we want to find the dimensions of the corresponding rectangle, we must
solve this system. Since calculations of areas and volumes are nonlinear in general,
situations involving geometric shapes often result in nonlinear systems of
equations.
To solve this system, we will start by dividing both sides of the second equation by ,
to obtain the following equivalent system.
If this second equation is satisfied, that means , which can be substituted into the
top equation to eliminate the variable .
The factor gives a solution of , and the factor gives a solution of .
Looking back at we see that if , then and if then .
There are two possible rectangles: One with width and length , and the other with
width and length .
Applications of Systems
A rectangle is drawn in the first quadrant with one side along the -axis, one side
along the -axis, the lower left corner at the origin, and upper right corner on the
graph of the equation . Denote this upper right vertex as . Find the coordinates of
the point if the area of the rectangle is .
Since are the coordinates of the upper right vertex, this tells us that and are both
positive. It also tells us that the distance from the -axis is , and the distance from the
-axis is . In other words, the height of the rectangle is just , and the width of the
rectangle is . In terms of and , the area is given by , giving us one equation
.
Since the upper right corner is on the graph of the line, we also know that .
This leaves us with the following system: This gives a system of nonlinear
equations
This bottom equation is already solved for , so the easiest way to eliminate a variable
would be to substitute it into the in the top equation.
We’ll solve this equation by completing the square.
If then , and if then .
There are two possibilities. One has coordinates and the other has coordinates
.
A right triangle has hypotenuse of length and area . Find the lengths of the two legs
of the triangle.
Since this is a right triangle, the Pythagorean Theorem tells us that . The area of a
triangle is given by which means , or equivalently .
This gives a system of nonlinear equations
A direct way to solve this system of equations would be to solve the bottom equation
for , giving , and substitute this into the top equation eliminating the variable. After
simplification that will yield a degree polynomial equation to solve for .
Instead of following that method, we will make use of a different algebraic
trick.
We know that there is a difference between and . If we multiply out we get . That
means if we add to , it becomes . To make use of that, we need to know the value of
so we can add it to the other side of our top equation. Notice that the bottom
equation of our system means that . If the bottom equation is satisfied, the top
equation can be rewritten as:
Taking square roots of both sides gives . That is, . Neither nor can be negative
(since they denote lengths of the sides of this triangle), this results in . Our system of
equations is equivalent to:
Now that we’ve been able to simplify the first equation, we will proceed with
substitution as mentioned above. If is satisfied, then the top equation gives:
The factor yields a solution of , and the factor gives a solution of . Using we find
that if then , and if then .
The two legs of the triangle have lengths and .
Suppose we have a box with square base, as illustrated below, constructed to have
volume and surface area . Call the side-lengths of the base as , and the height of the
box as .
Find the dimensions of the box.
We know that the volume of the box is given by (), giving the nonlinear equation .
The surface of the box consists of six rectangles. The top and bottom each have
area , and the four sides each have area . The full surface area of the box is
given by . This setup gives us a system of two nonlinear equations with two
unknowns:
If we want to find the dimensions of the box, we will have to solve this system of
equations.
As you have seen in the previous section, solving systems of nonlinear equations
involves finding a way to eliminate one of the variables by performing operations on
the two equations and/or using substutition. In the case of these equations,
notice that the variable only occurs in a single term in each equation. In the
top equation there is an term, while in the bottom equation there is an
term. These are not like terms, so we will need to deal with that. Let us
begin by multiplying both sides of the second equation by . This gives the
system
Since no solution has -coordinate equal to , this system is equivalent to the
original one. This modification has given that the variable appears in like
terms in boh equations. Substituting into the new bottom equation gives:
That is, if the equation is satisfied, then the bottom equation of the system is
equivalent to . This is a polynomial equation in the single variable, . (Notice that if
we had taken the original top equation , solved it for to obtain , and substituted that
into the original bottom equation, we would have arrived at this exact same
result.)
Notice that . That means is a solution to this cubic equation, and that is a factor of
the polynomial . By long-division we can find that . Since we see that . The zeroes of
this polymial are and . Since represents a length of the side of the box, the solution
is extraneous and should be dropped.
The only solution to the system has . Looking back at the first equation of the
system:
The solution is for . Since the question asks us to find the dimensions of the box, we
say that the box is . That is, it’s a cube with side length .
