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Worksheet 1
3.7 Optimization
PROBLEM 1 : Find two nonnegative numbers whose sum is 9 and so that the product
of one number and the square of the other number is a maximum.
Solution :
Let variables x and y represent two nonnegative numbers. The sum of the two
numbers is given to be
9=x+y,
so that
y=9-x.
We wish to MAXIMIZE the PRODUCT
P = x y2 .
However, before we differentiate the right-hand side, we will write it as a function
of x only. Substitute for y getting
P = x y2
= x ( 9-x)2 .
Now differentiate this equation using the product rule and chain rule, getting
P' = x (2) ( 9-x)(-1) + (1) ( 9-x)2
= ( 9-x) [ -2x + ( 9-x) ]
= ( 9-x) [ 9-3x ]
1
= ( 9-x) (3)[ 3-x ]
=0
For
x=9 or x=3 .
Note that since both x and y are nonnegative numbers and their sum is 9, it follows
that
. See the adjoining sign chart for P' .
If
x=3 and y=6 ,
then P= 108
is the largest possible product.
PROBLEM 2 : An open rectangular box with square base is to be made from 48 ft.2 of
material. What dimensions will result in a box with the largest possible volume ?
Solution :
Let variable x be the length of one edge of the square base and variable y the height
of the box.
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The total surface area of the box is given to be
48 = (area of base) + 4 (area of one side) = x2 + 4 (xy) ,
so that
4xy = 48 - x2
or
.
We wish to MAXIMIZE the total VOLUME of the box
V = (length) (width) (height) = (x) (x) (y) = x2 y .
However, before we differentiate the right-hand side, we will write it as a function
of x only. Substitute for y getting
V = x2 y
3
= 12x - (1/4)x3 .
Now differentiate this equation, getting
V' = 12 - (1/4)3x2
= 12 - (3/4)x2
= (3/4)(16 - x2 )
= (3/4)(4 - x)(4 + x)
=0
For
x=4 or x=-4 .
But
since variable x measures a distance and x > 0 . Since the base of the box
is square and there are 48 ft.2 of material, it follows that
adjoining sign chart for V' .
If
x=4 ft. and y=2 ft. ,
Then
V = 32 ft.3
. See the
is the largest possible volume of the box.
PROBLEM 3 : Consider all triangles formed by lines passing through the point
(8/9, 3) and both the x- and y-axes. Find the dimensions of the triangle with the
shortest hypotenuse.
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Solution :
Let variable x be the x-intercept and variable y the y-intercept of the line passing
throught the point (8/9, 3) .
Set up a relationship between x and y using similar
triangles.
One relationship is
,
so that
.
We wish to MINIMIZE the length of the HYPOTENUSE of the triangle
.
However, before we differentiate the right-hand side, we will write it as a function
of x only. Substitute for y getting
.
5
Now differentiate this equation using the chain rule and quotient rule, getting
(Factor a 2 out of the big brackets and simplify.)
=0,
so that (If
, then A=0 .)
.
By factoring out x , it follows that
,
so that (If AB= 0 , then A=0 or B=0 .)
x=0
(Impossible, since x> 8/9. Why ?) or
.
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Then
,
so that
(x-8/9)3 = 8 ,
x-8/9 = 2 ,
and
x = 26/9 .
See the adjoining sign chart for H' .
If
x = 26/9 and y=13/3 ,
then
is the shortest possible hypotenuse.
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PROBLEM 4 : Construct a window in the shape of a semi-circle over a rectangle. If the
distance around the outside of the window is 12 feet, what dimensions will result in
the rectangle having largest possible area ?
Solution :
Let variable x be the width and variable y the length of the rectangular portion of the
window.
𝑋
The semi-circular portion of the window has length:𝜋( )
2
The perimeter (distance around outside only) of the window is given to be
𝑥
12 = 𝑥 + 2𝑦 + 𝜋( )
2
.
We wish to MAXIMIZE the total AREA of the RECTANGLE
A = (width) (length) = x y .
However, before we differentiate the right-hand side, we will write it as a function
of x only. Substitute for y getting
A=xy
8
.
Now differentiate this equation, getting
=0
For
,
i.e.,
.
Since variable x measures distance,
. In addition, x is largest when y = 0 and the
window is in the shape of a semi-circle. Thus,
adjoining sign chart for A' .
If
ft. and y=3 ft. ,
ft.2
Then
9
is the largest possible area of the rectangle.
(Why ?). See the
Problem 5 : A rectangular page is to contain 30 square inches of print . The margins of each side are 1
inch.Find the dimensions of the page such that the least amount of paper is used . (example 3 p.220)
Problem 6 : Because the box has a square base, its volume is
V = x2h.
Primary equation
This equation is called the primary equation because it gives a formula for the quantity to be optimized.
The surface area of the box is
S = (area of base) + (area of four sides)
S = x2 + 4xh = 108.
Secondary equation
Because V is to be maximized, you want to write V as a function of just one variable.
To do this, you can solve the equation x2 + 4xh = 108 for h in terms of x to obtain h = (108 – x2)/(4x).
Substituting into the primary equation produces
Before finding which x-value will yield a maximum value of V, you should determine the feasible domain.
That is, what values of x make sense in this problem?
You know that V ≥ 0. You also know that x must be nonnegative and that the area of the base (A = x2) is
at most 108.
So, the feasible domain is
To maximize V, find the critical numbers of the volume
function on the interval
So, the critical numbers are x = ±6.
You do not need to consider x = –6 because it is outside the
domain.
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Evaluating V at the critical number x = 6 and at the endpoints
of the domain produces V(0) = 0, V(6) = 108, and
So, V is maximum when x = 6 and the dimensions of the box
are 6 х 6 х 3 inches.
Problem7:
The distance between the point (0, 2) and a point (x, y) on the graph of y = 4 – x2 is
given by :
Using the secondary equation y = 4 – x2, you can rewrite the primary equation as
Because d is smallest when the expression inside the radical is smallest, you need only find the critical
numbers of f(x) = x4 – 3x2 + 4.
Note that the domain of f is the entire real line. So, there are no endpoints of the domain to consider
Moreover, setting f'(x) equal to 0 yields :
The First Derivative Test verifies that x = 0 yields a relative maximum, whereas the other solutions
yield a minimum distance.
So, the closest points are
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