Let’s begin by letting x and y re present the numbers we
seek. Next, let’s play
a bit as we did in the previous example. Try x = 9 and y = 1. The difference of
these two numbers is certainly 8. The sum of the squares of these two numbers is
S = 92 + 12 = 82. Let’s put this result in tabular form.
For a second guess, select x = 8 and y = 0. The difference
is x−y = 8−0 = 8, but
this time the sum of the squares is S = 82 + 02 = 64. For a third guess, try x
= 7 and
y = −1. Again, the difference is x − y = 7 − (−1) = 8, but the sum of the
now S = 72 + (−1)2 = 50. Let’s add these results to our table.
Thus far, the pair that minimizes the sum of the squares
is x = 7 and y = −1.
However, could there be another pair with a difference of 8 and the sum of the
is smaller than 50? Experiment further to see if you can best the current
Let’s try an analytical approach. Our first constraint is
the fact that the difference
of the two numbers must equal 8. This is easily expressed as
x − y = 8. (13)
Next, we’re asked to minimize the sum of the squares of
the two numbers. This requires
that we find a formula for the sum of the squares. Let S represent the sum of
of x and y. Thus,
Compare S = 2y2+16y+64 with the general quadratic S =
ay2+by+c and note that
a = 2 and b = 16. Thus, the plot of S versus y will be a parabola that opens
(see Figure 4) and the minimum value of S will occur at the vertex. The
of the vertex is found with
Figure 4. Plotting the sum of the
squares S versus y. The minimum S,
32, occurs at the vertex, (−4, 32).
Thus, the first number we seek is y = −4. We can find the
second number by substituting
y = −4 in equation (15).
x = y + 8 = (−4) + 8 = 4.
Hence, the numbers we seek are x = 4 and y = −4. Note that
the difference of these two
numbers is x−y = 4−(−4) = 8 and the sum of their squares is S = (4)2+(−4)2 =
which is smaller than the best result found in our tabular experiment above.
our work show that this is the smallest possible value of S.
Alternatively, you can find S by substituting y = −4 in
equation (16). We’ll leave
it to our readers to verify that this also gives a minimum value of S = 32.
Let’s look at another application.
Example 17. Mary wants to fence a rectangular garden to
keep the deer from
eating her fruit and vegetables. One side of her garden abuts her shed wall so
not need to fence that side. However, she also wants to use material to separate
rectangular garden in two sections (see Figure 5). She can afford to buy 80
of fencing to use for the perimeter and the section dividing the rectangular
What dimensions will maximize the total area of the rectangular garden?
Figure 5. Mary’s rectangular garden
needs fencing on three sides and also for
the fence to divide the garden.
Again, before we take an algebraic approach , let’s just
experiment. Note that we’ve
labeled the width with the letter x and the height with the letter y in our
sketch of the
garden in Figure 5.
There is a total of 80 feet of fence material. Suppose
that we let y = 5 ft. Because
there are three sides of length y = 5 ft, we’ve used 15 feet of material. That
feet of material which will be used to fence the width of the garden. That is,
is x = 65 ft. Thus, the dimensions of the garden are x = 65 ft by y = 5 ft. The
equals the product of these two measures, so A = 325 ft2. Let’s put this result
Suppose instead that we let the height be y = 10 ft.
Again, there are three sections
with this length, so this will take 30 ft of material. That leaves 50 ft of
material, so the
width x = 50 ft. The area is the product of these two
measures, so A = 500 ft2. As
a third experiment, let the height y = 15 ft. Subtracting three of these lengths
80 ft, we see that the width x = 35 ft. The area is the product of these
A = 525 ft2. Let’s add these last two number experiments to our table.
At this point, the last set of dimensions yields the
maximum area, but is it possible
that another choice of x and y will yield a larger area? Experiment further with
of your choice to see if you can find dimensions that will yield an area larger
current maximum in the table, namely 525 ft2.
Let’s now call on what we’ve learned in this section to
attack this model. First,
we’re constrained by the amount of material we have for the job, a total of 80
fencing. This constraint requires that 3 times the height of the garden, added
width of the garden, should equal the available amount of fencing material. In
x + 3y = 80. (18)
We’re asked to maximize the area, so we focus our efforts
on finding a formula for the
area of the rectangular garden. Because the area A of the rectangular garden is
product of the width and the height,
A = xy. (19)
We now have a formula for the area of the rectangular
garden, but unfortunately we
have the area A as a function of two variables. We need to eliminate one or the
of these variables. This is easily done by solving equation (18) for x.
x = 80 − 3y (20)
Next, substitute this result in equation (19) to get
Note that we have expressed the area A as a function of a
single variable y. Also, the
function defined by equation (21) is quadratic. Compare A = −3y2 + 80y with the
general form A = ay2 + by + c and note that a = −3 and b = 80 (we have no need
of the fact that c = 0). Therefore, if we plot A versus y, the graph is a
opens downward (see Figure 6), so the maximum value of A will occur at the
The y-coordinate of the vertex is found with
To find the width of the rectangular garden, substitute y
= 40/3 into equation (20)
and solve for x.
Thus, the width of the rectangular garden is 40 ft. We can
find the area of the garden
by multiplying the width and the height.
Note that the resulting area,
is only slightly bigger the last tabular
found with our numerical experiments .
You can also find the area of the rectangular region by
substituting y = 40/3 into
equation (21). We’ll leave it to our readers to check that this provides the
measure for the area. You will also notice that the second coordinate of the
Figure 6 is the maximum area A = 1600/3 ft2.
Figure 6. The maximum area, A =
1600/3 ft2, occurs at the vertex of the
parabola, (40/3, 1600/3).