Solutions to Maximum/Minimum Problems
SOLUTIONS TO MAXIMUM/MINIMUM PROBLEMS
SOLUTION 15 : Let variable r be the radius of the circular base and variable h the height of the inscribed cone as
shown in the two-dimensional side view.
It is given that the circle's radius is 2. Find a relationship between r and h . Let variable z be the height of the small
right triangle.
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Solutions to Maximum/Minimum Problems
By the Pythagorean Theorem it follows that
r 2 + z 2 = 22
so that
z2 = 4 - r2
or
.
Thus the height of the inscribed cone is
.
We wish to MINIMIZE the total VOLUME of the CONE
.
Now differentiate this equation using the product rule and the chain rule, getting
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Solutions to Maximum/Minimum Problems
(Factor out
, get a common denominator, and simplify fractions.)
(Factor out (r) .)
=0,
so that (If AB = 0 , then A=0 or B=0 .)
r=0
or
,
i.e., (If
, then A=0 .)
.
Then (Isolate the square root term.)
,
,
(Square both sides of this equation.)
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Solutions to Maximum/Minimum Problems
,
16 ( 4 - r 2 ) = 9 r 4 - 48 r 2 + 64 ,
64 - 16 r 2 = 9 r 4 - 48 r 2 + 64 ,
32 r 2 - 9 r 4 = 0 ,
r 2 ( 32 - 9 r 2 ) = 0 ,
so that
r=0
or
32 - 9 r 2 = 0 ,
r 2 = 32/9 ,
or
.
But
since variable r measures a distance and
. See the adjoining sign chart for V' .
If
and
,
then
is the largest possible volume for the inscribed cone.
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Solutions to Maximum/Minimum Problems
SOLUTION 16 : Write the area of the given isosceles triangle as a function of . Let variable x be the length of the
base and variable y the height of the triangle, and consider angle
. Write each of x and y as functions of .
It follows from basic trigonometry that
so that
(Equation 1 )
,
and
so that
(Equation 2 )
We wish to MAXIMIZE the AREA of the isosceles triangle
A = (1/2) (length of base) (height) = (1/2) xy .
Before we differentiate, use Equations 1 and 2 to rewrite the right-hand side as a function of
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only. Then
Solutions to Maximum/Minimum Problems
A = (1/2) xy
.
Now differentiate this equation using the product rule and chain rule, getting
(Factor out (9/2) and simplify the expression.)
=0,
so that
and
.
It follows algebraically (Why ?) that
so that from basic trigonometry we get
or
,
or
.
and hence
Because
measures an angle in a triangle, it is logical to assume that
. Thus,
sign chart for A' .
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. See the adjoining
Solutions to Maximum/Minimum Problems
If
radians = 90 degrees,
then
A = 9/2
is the largest possible area for the triangle.
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SOLUTION 17 : We need to determine a general SLOPE EQUATION for tangent lines.
This means that we need the first derivative of y . Differentiate
using the quotient rule, getting
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Solutions to Maximum/Minimum Problems
.
We wish to MAXIMIZE and MINIMIZE the SLOPE equation
.
Now differentiate this equation using the quotient rule and chain rule, getting
(Factor out -12 and (x 2 +3) from the numerator and simplify the expression.)
(Divide out a factor of (x 2 +3) .)
=0,
so that (If
, then A = 0 .)
-36 ( 1 - x ) ( 1 + x ) = 0
and (If AB = 0 , then A = 0 or B = 0 .)
x=1 or x=-1 .
See the adjoining sign chart for S' .
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Solutions to Maximum/Minimum Problems
If
x=-1 and y = 3/2 ,
then
S= 3/4
is the largest possible slope for this graph. The corresponding tangent line is
y - 3/2 = 3/4( x - (-1) )
or
y = (3/4)x + (9/4) .
If
x=1 and y = 3/2 ,
then
S= -3/4
is the smallest possible slope for this graph. The corresponding tangent line is
y - 3/2 = -3/4( x - 1 )
or
y = (-3/4)x + (9/4) .
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SOLUTION 18 : Let variable L be the length of the ladder resting on the top of the fence and touching the wall behind
it. Let variables x and y be the lengths as shown in the diagram.
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Solutions to Maximum/Minimum Problems
Write L as a function of x . First find a relationship between y and x using similar triangles. For example,
so that
.
We wish to MINIMIZE the LENGTH of the ladder
.
Before we differentiate, rewrite the right-hand side as a function of x only. Then
.
Now differentiate this equation using the chain rule and quotient rule, getting
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Solutions to Maximum/Minimum Problems
(Factor out 64x and (x-3) from the numerator of the fraction inside the brackets.)
(Divide out a factor of (x-3) and simplify the entire expression.)
(Factor out 2x from the numerator.)
=0,
so that (If
, then A = 0 .)
.
Then (If AB = 0 , then A = 0 or B = 0 .)
x=0
or
,
so that
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Solutions to Maximum/Minimum Problems
,
(x-3) 3 = 192 ,
,
and
.
Note that x > 3 . See the adjoining sign chart for L' .
If
ft. and
ft. ,
then
ft.
is the length of the shortest possible ladder.
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SOLUTION 19 :
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Solutions to Maximum/Minimum Problems
Let variable S be the sum of the squares of the distances between (0, 0) and (x, 0) ,
,
and between (3, 2) and (x, 0) ,
.
We wish to MINIMIZE the SUM of the squares of the distances
S = x 2 + ( x 2 -6x +13 ) = 2x 2 -6x +13 .
Now differentiate, getting
S' = 4x -6
= 4(x - 3/2)
=0
for
x= 3/2 .
See the adjoining sign chart for S' .
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Solutions to Maximum/Minimum Problems
If
x = 3/2 ,
then
S = 17/2
is the smallest sum.
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SOLUTION 20 :
Assume that the two cars travel at the following rates :
CAR A : 60 mph
CAR B : 90 mph
Let variable x be the distance car A travels in t hours, and variable y the distance car B travels in t hours. Let variable
L be the distance between cars A and B after t hours.
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Solutions to Maximum/Minimum Problems
Thus, by the Pythagorean Theorem distance L is
.
Before we differentiate, we will rewrite the right-hand side as a function of t only. Recall that if travel is at a
CONSTANT rate then
(distance traveled) = (rate of travel) (time elapsed) .
Thus, for car A the distance traveled after t hours is
(Equation 1 )
x = 60 t ,
and for car B the distance traveled after t hours is
(Equation 2 )
y = 90 t .
Use Equations 1 and 2 to rewrite the equation for L as a function of t only. Thus, we wish to MINIMIZE the
DISTANCE between the two cars
.
Differentiate, using the chain rule, getting
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Solutions to Maximum/Minimum Problems
=0
so that (If
, then A = 0 .)
23,400 t - 5400 = 0 ,
and
.
See the adjoining sign chart for L' .
If
hrs. = 13.8 min. ,
then
mi. ,
mi. ,
and
mi.
is the shortest possible distance between the cars.
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SOLUTION 21 : Let variable L represent the length of the crease and let variables x and y be as shown in the diagram.
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Solutions to Maximum/Minimum Problems
We wish to write L as a function of x . Introduce variable w as shown in the following diagram.
It follows from the Pythagorean Theorem that
w2 + (6-x) 2 = x 2 ,
so that
w2 = x 2 - (x 2 - 12x + 36) = 12x - 36
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Solutions to Maximum/Minimum Problems
and
.
Find a relationship between x and y . The total area of the paper can be computed from the areas of three right
triangles, two of which are exactly the same dimensions, and one rhombus. In particular
72 = (total area of paper)
= (area of small triangle) + 2(area of large triangle) + (area of rhombus)
= (1/2)(length of base)(height) + 2(1/2)(length of base)(height) + (average height)(length of base)
,
i.e.,
.
Solve this equation for y . Then
,
,
,
and
.
We wish to MINIMIZE the LENGTH of the crease
.
Before we differentiate, rewrite the right-hand side as a function of x only. Then
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Solutions to Maximum/Minimum Problems
.
Now differentiate this equation using the chain rule and quotient rule, getting
(Factor out x from the numerator.)
=0,
so that (If
, then A = 0 .)
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Solutions to Maximum/Minimum Problems
.
Thus, (If AB = 0 , then A = 0 or B = 0 .)
x=0
or
,
,
-2 (x-3) 2 = 3x - 18 ,
-2 (x 2 - 6x + 9) = 3x - 18 ,
-2 x 2 + 12x -18 = 3x - 18 ,
-2 x 2 + 9x = 0 ,
x ( -2x + 9 ) = 0 ,
so that (If AB = 0 , then A = 0 or B = 0 .)
x=0
or
( -2x + 9 ) = 0 ,
i.e.,
x = 9/2 .
Note that since the paper is 6 inches wide, it follows that
. See the adjoining sign chart for L' .
If
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Solutions to Maximum/Minimum Problems
x = 9/2 in. and
in.
in. ,
then
in.
in.
is the length of the shortest possible crease.
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About this document ...
Duane Kouba
1998-06-17
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