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.

By the Pythagorean Theorem it follows that

r2 + z2 = 22

so that

z2 = 4 - r2

or

$ z = \sqrt{ 4 - r^2 } $ .

Thus the height of the inscribed cone is

$ h = 2 + z = 2 + \sqrt{ 4 - r^2 } $ .

We wish to MINIMIZE the total VOLUME of the CONE

$ V = (1/3) \pi r^2 h $ .

However, before we differentiate the right-hand side, we will write it as a function of r only. Substitute for h getting

$ V = (1/3) \pi r^2 h $

$ = (1/3) \pi r^2 ( 2 + \sqrt{ 4 - r^2 } ) $ .

Now differentiate this equation using the product rule and the chain rule, getting

$ V' = (1/3) \pi r^2 \Big\{ 0 + (1/2)( 4 - r^2 )^{-1/2} (-2r) \Big\} + (1/3) \pi 2 r \big( 2 + \sqrt{ 4 - r^2 } \big) $

(Factor out $ ( \pi/3 ) $ , get a common denominator, and simplify fractions.)

$ = \displaystyle{ \pi \over 3 } \Big\{ \displaystyle{ -r^3 \over \sqrt{ 4 - r^2 } }
+ 2 r \big( 2 + \sqrt{ 4 - r^2 } \big) \Big\} $

$ = \displaystyle{ \pi \over 3 } \Big\{ \displaystyle{ -r^3 \over \sqrt{ 4 - r^2...
... - r^2 } \big) \displaystyle{ \sqrt{ 4 - r^2 } \over \sqrt{ 4 - r^2 } } \Big\} $

$ = \displaystyle{ \pi \over 3 } \Big\{
\displaystyle{ -r^3 + 2 r \big( 2 + \sqrt{ 4 - r^2 } \big) \sqrt{ 4 - r^2 } \over \sqrt{ 4 - r^2 } } \Big\} $

$ = \displaystyle{ \pi \over 3 } \Big\{
\displaystyle{ -r^3 + 2 r \big( 2 \sqrt{ 4 - r^2 } + (4-r^2) \big) \over \sqrt{ 4 - r^2 } } \Big\} $

(Factor out (r) .)

$ = \displaystyle{ \pi \over 3 } (r) \Big\{
\displaystyle{ -r^2 + 2 \big( 2 \sqrt{ 4 - r^2 } + (4-r^2) \big) \over \sqrt{ 4 - r^2 } } \Big\} $

= 0 ,

so that (If AB = 0 , then A=0 or B=0 .)

r = 0

or

$ \displaystyle{ -r^2 + 2 ( 2 \sqrt{ 4 - r^2 } + (4-r^2) ) \over \sqrt{ 4 - r^2 } } = 0 $ ,

i.e., (If $ \displaystyle{ A \over B } = 0 $ , then A=0 .)

$ -r^2 + 2 ( 2 \sqrt{ 4 - r^2 } + (4-r^2) ) = 0 $ .

Then (Isolate the square root term.)

$ 4 \sqrt{ 4 - r^2 } + 8 - 2 r^2 = r^2 $ ,

$ 4 \sqrt{ 4 - r^2 } = 3 r^2 - 8 $ ,

(Square both sides of this equation.)

$ 16 ( \sqrt{ 4 - r^2 } )^2 = (3 r^2 - 8)^2 $ ,

16 ( 4 - r2 ) = 9 r4 - 48 r2 + 64 ,

64 - 16 r2 = 9 r4 - 48 r2 + 64 ,

32 r2 - 9 r4 = 0 ,

r2 ( 32 - 9 r2 ) = 0 ,

so that

r = 0

or

32 - 9 r2 = 0 ,

r2 = 32/9 ,

or

$ r = \pm \sqrt{ 32/9 } = \pm 4 \sqrt{ 2 }/3 \approx \pm 1.89 $ .

But $ r \ne - 4 \sqrt{ 2 }/3 $ since variable r measures a distance and $ 0 \le r \le 2 $ . See the adjoining sign chart for V' .

If

$ r= 4 \sqrt{ 2 }/3 \approx 1.89 $ and $ h= 8/3 \approx 2.67 $ ,

then

$ V \approx 9.93 $

is the largest possible volume for the inscribed cone.

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SOLUTION 16 : Write the area of the given isosceles triangle as a function of $ \theta $ . Let variable x be the length of the base and variable y the height of the triangle, and consider angle $ \theta/2 $ . Write each of x and y as functions of $ \theta $ .

It follows from basic trigonometry that

$ \cos \Big( \displaystyle{ \theta \over 2 } \Big) = \displaystyle{ y \over 3 } $

so that

(Equation 1 )

$ y = 3 \cos \Big( \displaystyle{ \theta \over 2 } \Big) $ ,

and

$ \sin \Big( \displaystyle{ \theta \over 2 } \Big) = \displaystyle{ \displaystyle{ x \over 2 } \over 3 }
= \displaystyle{ x \over 6 } $

so that

(Equation 2 )

$ x = 6 \sin \Big( \displaystyle{ \theta \over 2 } \Big) $

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 $ \theta $ only. Then

A = (1/2) xy

$ = (1/2) ( 6 \sin \Big( \displaystyle{ \theta \over 2 } \Big) ) ( 3 \cos \Big( \displaystyle{ \theta \over 2 } \Big) ) $

$ = 9 \sin \Big( \displaystyle{ \theta \over 2 } \Big) \cos \Big( \displaystyle{ \theta \over 2 } \Big) $ .

Now differentiate this equation using the product rule and chain rule, getting

$ A' = 9 \sin \Big( \displaystyle{ \theta \over 2 } \Big)
\big\{ -\sin \Big( \...
...over 2 } \Big) (1/2) \big\}
\cos \Big( \displaystyle{ \theta \over 2 } \Big) $

(Factor out (9/2) and simplify the expression.)

$ = \displaystyle{ 9 \over 2 }
\big\{ \cos^2 \Big( \displaystyle{ \theta \over 2 } \Big)
- \sin^2 \Big( \displaystyle{ \theta \over 2 } \Big) \big\} $

= 0 ,

so that

$ \cos^2 \Big( \displaystyle{ \theta \over 2 } \Big)
- \sin^2 \Big( \displaystyle{ \theta \over 2 } \Big) = 0 $

and

$ \cos^2 \Big( \displaystyle{ \theta \over 2 } \Big)
= \sin^2 \Big( \displaystyle{ \theta \over 2 } \Big) $ .

It follows algebraically (Why ?) that

$ \cos \Big( \displaystyle{ \theta \over 2 } \Big)
= \pm \sin \Big( \displaystyle{ \theta \over 2 } \Big) $

so that from basic trigonometry we get

$ \displaystyle{ \theta \over 2 } = \displaystyle{ \pi \over 4 } $ or $ \displaystyle{ \theta \over 2 } = \displaystyle{ 3 \pi \over 4 } $ ,

and hence

$ \theta = \displaystyle{ \pi \over 2 } $ or $ \theta = \displaystyle{ 3 \pi \over 2 } $ .

Because $ \theta $ measures an angle in a triangle, it is logical to assume that $ 0 \le \theta \le \pi $ . Thus, $ \theta \ne \displaystyle{ 3 \pi \over 2 } $ . See the adjoining sign chart for A' .

If

$ \theta = \displaystyle{ \pi \over 2 } $ 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 $ y = \displaystyle{ 6 \over x^2+3 } $ using the quotient rule, getting

$ y' = \displaystyle{ (x^2 + 3) (0) - (6) ( 2x ) \over (x^2 + 3)^2 } $

$ = \displaystyle{ -12x \over (x^2 + 3)^2 } $ .

We wish to MAXIMIZE and MINIMIZE the SLOPE equation

$ S = \displaystyle{ -12x \over (x^2 + 3)^2 } $ .

Now differentiate this equation using the quotient rule and chain rule, getting

$ S' = \displaystyle{ (x^2 + 3)^2 (-12) -(-12x)2 (x^2 + 3)(2x) \over (x^2 + 3)^4 } $

(Factor out -12 and (x2+3) from the numerator and simplify the expression.)

$ = \displaystyle{ -12 (x^2 + 3) [ (x^2 + 3) - 4x^2 ]\over (x^2 + 3)^4 } $

(Divide out a factor of (x2+3) .)

$ = \displaystyle{ -12 ( 3 - 3x^2 ) \over (x^2 + 3)^3 } $

$ = \displaystyle{ -12 ( 3 ) ( 1 - x^2 ) \over (x^2 + 3)^3 } $

$ = \displaystyle{ -36 ( 1 - x ) ( 1 + x ) \over (x^2 + 3)^3 } $

= 0 ,

so that (If $ \displaystyle{ A \over B } = 0 $ , 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' .

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.

Write L as a function of x . First find a relationship between y and x using similar triangles. For example,

$ \displaystyle{ y \over x } = \displaystyle{ 8 \over x-3 } $

so that

$ y = \displaystyle{ 8x \over x-3 } $ .

We wish to MINIMIZE the LENGTH of the ladder

$ L = \sqrt{ x^2 + y^2 } $ .

Before we differentiate, rewrite the right-hand side as a function of x only. Then

$ L = \sqrt{ x^2 + y^2 } $

$ = \sqrt{ x^2 + \Big( \displaystyle{ 8x \over x-3 } \Big)^2 } $

$ = \sqrt{ x^2 + \displaystyle{ 64x^2 \over (x-3)^2 } } $ .

Now differentiate this equation using the chain rule and quotient rule, getting

$ L' = (1/2) \Big( x^2 + \displaystyle{ 64x^2 \over (x-3)^2 } \Big)^{-1/2}
\Big\{ 2x + \displaystyle{ (x-3)^2(128x) - (64x^2)2(x-3) \over (x-3)^4 } \Big\} $

(Factor out 64x and (x-3) from the numerator of the fraction inside the brackets.)

$ = (1/2) \Big( x^2 + \displaystyle{ 64x^2 \over (x-3)^2 } \Big)^{-1/2}
\Big\{ 2x + \displaystyle{64x(x-3) [ 2(x-3) - 2x ] \over (x-3)^4 } \Big\} $

(Divide out a factor of (x-3) and simplify the entire expression.)

$ = \displaystyle{ 2x + \displaystyle{64x (-6) \over (x-3)^3 } \over
2 \sqrt{ x^2 + \displaystyle{ 64x^2 \over (x-3)^2 } } } $

(Factor out 2x from the numerator.)

$ = \displaystyle{ 2x \Big\{ 1 + \displaystyle{ -192 \over (x-3)^3 } \Big\} \over
2 \sqrt{ x^2 + \displaystyle{ 64x^2 \over (x-3)^2 } } } $

$ = \displaystyle{ x \Big\{ 1 - \displaystyle{ 192 \over (x-3)^3 } \Big\} \over
\sqrt{ x^2 + \displaystyle{ 64x^2 \over (x-3)^2 } } } $

= 0 ,

so that (If $ \displaystyle{ A \over B } = 0 $ , then A = 0 .)

$ x \Big\{ 1 - \displaystyle{ 192 \over (x-3)^3 } \Big\} = 0 $ .

Then (If AB = 0 , then A = 0 or B = 0 .)

x=0

(This is impossible since x>3.)

or

$ 1 - \displaystyle{ 192 \over (x-3)^3 } = 0 $ ,

so that

$ 1 = \displaystyle{ 192 \over (x-3)^3 } $ ,

(x-3)3 = 192 ,

$ x-3 = 192 ^{1/3} \approx 5.77 $ ,

and

$ x \approx 8.77 $ .

Note that x > 3 . See the adjoining sign chart for L' .

If

$ x \approx 8.77 $ ft. and y is approximately 12.16 ft. ,

then

L is approximately 14.99 ft.

and is the length of the shortest possible ladder.

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SOLUTION 19 :

Let variable S be the sum of the squares of the distances between (0, 0) and (x, 0) ,

$ \big( \sqrt{ (x-0)^2 + (0-0)^2 } \big)^2 = x^2 $ ,

and between (3, 2) and (x, 0) ,

$ \big( \sqrt{ (x-3)^2 + (0-2)^2 } \big)^2 = (x-3)^2 + 4 = x^2 -6x +13 $ .

We wish to MINIMIZE the SUM of the squares of the distances

S = x2 + ( x2 -6x +13 ) = 2x2 -6x +13 .

Now differentiate, getting

S' = 4x -6

= 4(x - 3/2)

= 0

for

x= 3/2 .

See the adjoining sign chart for S' .

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 : North at 60 mph

CAR B : West at 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.

Thus, by the Pythagorean Theorem distance L is

$ L = \sqrt{ x^2 + (30 - y)^2 } $ .

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

$ L = \sqrt{ x^2 + (30 - y)^2 } $

$ = \sqrt{ ( 60 t )^2 + (30 - 90 t )^2 } $

$ = \sqrt{ 3600 t^2 + (30 - 90 t )^2 } $ .

Differentiate, using the chain rule, getting

$ L' = (1/2) \big( 3600 t^2 + (30 - 90 t )^2 \big)^{-1/2} \{ 7200 t + 2 (30 - 90 t ) (-90) \} $

$ = \displaystyle{ 23,400 t - 5400 \over 2 \sqrt{ 3600 t^2 + (30 - 90 t )^2 } } $

= 0

so that (If $ \displaystyle{ A \over B } = 0 $ , then A = 0 .)

23,400 t - 5400 = 0 ,

and

$ t \approx 0.23 $ .

See the adjoining sign chart for L' . (Please note that there are TWO ERRORS in the adjoining sign chart. The correct value for y is approximately 9.23 miles, NOT 20.77 miles. The correct value for L is approximately 16.64 miles, NOT 24.96 miles.)

If

$ t \approx 0.23 $ hrs. = 13.8 min. ,

then

$ x \approx 13.85 $ mi. , $ y \approx 20.77 $ mi. ,

and

$ L \approx 24.96 $ 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.

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 = x2 ,

so that

w2 = x2 - (x2 - 12x + 36) = 12x - 36

and

$ w = \sqrt{ 12x - 36 } $ .

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 trapezoid. In particular

72 = (total area of paper)

= (area of small triangle) + 2(area of large triangle) + (area of trapezoid)

= (1/2)(length of base)(height) + 2(1/2)(length of base)(height) + (average height)(length of base)

$ = (1/2)(6-x)( \sqrt{ 12x - 36 } ) + (x)(y) + (1/2) \big\{ (12-y) + (12 - \sqrt{ 12x - 36 } \big\} (6) $

$ = ( 3-(1/2)x ) \sqrt{ 12x - 36 } + xy + 3 \big\{ 24 - y - \sqrt{ 12x - 36 } \big\} $

$ = 3 \sqrt{ 12x - 36 } - (1/2)x \sqrt{ 12x - 36 } + xy + 72 - 3y - 3 \sqrt{ 12x - 36 } $

$ = - (1/2)x \sqrt{ 12x - 36 } + xy + 72 - 3y $ ,

i.e.,

$ 72 = - (1/2)x \sqrt{ 12x - 36 } + xy + 72 - 3y $ .

Solve this equation for y . Then

$ 72 = - (1/2)x \sqrt{ 12x - 36 } + xy + 72 - 3y $ ,

$ 0 = - (1/2)x \sqrt{ 12x - 36 } + ( x - 3 ) y $ ,

$ ( x - 3 ) y = (1/2)x \sqrt{ 12x - 36 } $ ,

and

$ y = \displaystyle{ x \sqrt{ 12x - 36 } \over 2 (x - 3) } $ .

We wish to MINIMIZE the LENGTH of the crease

$ L = \sqrt{ x^2 + y^2 } $ .

Before we differentiate, rewrite the right-hand side as a function of x only. Then

$ L = \sqrt{ x^2 + y^2 } $

$ = \sqrt{ x^2 + \Big( \displaystyle{ x \sqrt{ 12x - 36 } \over 2 (x - 3) } \Big)^2 } $

$ = \sqrt{ x^2 + \displaystyle{ x^2 (12x - 36) \over 4 (x-3)^2 } } $

$ = \sqrt{ x^2 + \displaystyle{ x^2 12(x - 3) \over 4 (x-3)^2 } } $

$ = \sqrt{ x^2 + \displaystyle{ 3x^2 \over x-3 } } $ .

Now differentiate this equation using the chain rule and quotient rule, getting

$ L' = (1/2) \Big( x^2 + \displaystyle{ 3x^2 \over x-3 } \Big)^{-1/2}
\Big\{ 2x + \displaystyle{ (x-3) (6x) - (3x^2) (1) \over (x-3)^2 } \Big\} $

$ =
\displaystyle{ 2x + \displaystyle{ 3x^2 - 18x \over (x-3)^2 } \over
2 \sqrt{ x^2 + \displaystyle{ 3x^2 \over x-3 } } } $

$ =
\displaystyle{ 2x + \displaystyle{ x (3x - 18) \over (x-3)^2 } \over
2 \sqrt{ x^2 + \displaystyle{ 3x^2 \over x-3 } } } $

(Factor out x from the numerator.)

$ =
\displaystyle{ ( x ) \Big\{ 2 + \displaystyle{ 3x - 18 \over (x-3)^2 } \Big\} \over
2 \sqrt{ x^2 + \displaystyle{ 3x^2 \over x-3 } } } $

= 0 ,

so that (If $ \displaystyle{ A \over B } = 0 $ , then A = 0 .)

$ ( x ) \Big\{ 2 + \displaystyle{ 3x - 18 \over (x-3)^2 } \Big\} = 0 $ .

Thus, (If AB = 0 , then A = 0 or B = 0 .)

x = 0

or

$ 2 + \displaystyle{ 3x - 18 \over (x-3)^2 } = 0 $ ,

$ -2 = \displaystyle{ 3x - 18 \over (x-3)^2 } $ ,

-2 (x-3)2 = 3x - 18 ,

-2 (x2 - 6x + 9) = 3x - 18 ,

-2 x2 + 12x -18 = 3x - 18 ,

-2 x2 + 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 $ 3 < x \le 6 $ . See the adjoining sign chart for L' .

If

x = 9/2 in. and $ y = 9/\sqrt{ 2 } $ in. $ \approx 6.36 $ in. ,

then

$ L = 9\sqrt{ 3 }/2 $ in. $ \approx 7.79 $ in.

is the length of the shortest possible crease.

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Duane Kouba
1998-06-17