SOLUTIONS TO MAXIMUM/MINIMUM PROBLEMS


SOLUTION 8 : Let variable r be the radius of the circular base and variable h the height of the cylinder.

The total volume of the cylinder is given to be

$ 20 \pi = $ (area of base) (height) $ = ( \pi r^2 ) h $ ,

so that

$ h = \displaystyle{ 20 \pi \over \pi r^2 } $

$ = \displaystyle{ 20 \over r^2 } $ .

We wish to MINIMIZE the total COST of construction of the cylinder

C = (total cost of bottom) + (total cost of top) + (total cost of side)

= (unit cost of bottom)(area of bottom) + (unit cost of top)(area of top) + (unit cost of side) (area of side)

$ = \$10 (\pi r^2) + \$10 (\pi r^2) + \$8 (2 \pi r h) $

(For convenience drop the $ \$ $ signs until the end of the problem.)

$ = 20 \pi r^2 + 16 \pi r h $ .

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

$ C = 20 \pi r^2 + 16 \pi r h $

$ = 20 \pi r^2 + 16 \pi r \Big( \displaystyle{ 20 \over r^2 } \Big) $

$ = 20 \pi r^2 + \displaystyle{ 320 \pi \over r } $ .

Now differentiate this equation, getting

$ C' = 40 \pi r + 320 \pi \Big\{ \displaystyle{ -1 \over r^2 } \Big\} $

$ = 40 \pi r - \displaystyle{ 320 \pi \over r^2 } $

(Get a common denominator and combine fractions.)

$ = 40 \pi r \Big\{ \displaystyle{ r^2 \over r^2 } \Big\} - \displaystyle{ 320 \pi \over r^2 } $

$ = \displaystyle{ 40 \pi r^3 - 320 \pi \over r^2 } $

$ = \displaystyle{ 40 \pi (r^3 - 8) \over r^2 } $

= 0 ,

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

$ 40 \pi (r^3 - 8) = 0 $ ,

r3 = 8 ,

or

r = 2 .

Since variable r measures a distance, it must satisfy r > 0 . See the adjoining sign chart for C' .

If

r=2 m. and h=5 m. ,

then

$ C = \$ 240 \pi \approx \$ 754 $

is the least possible cost of construction.

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SOLUTION 9 : Let variable x be the distance denoted in the given diagram.

Assume that you travel at the following rates :

SWIM : 2 mph

WALK : 3 mph .

Recall that if travel is at a CONSTANT rate of speed, then

(distance traveled) = (rate of travel) (time elapsed)

or

D = R T ,

so that time elapsed is

$ T = \displaystyle{ D \over R } $ .

We wish to MINIMIZE the total TIME elapsed

T = (swim time) + (walk time)

= (swim distance)/(swim rate) + (walk distance)/(walk rate)

$ = \displaystyle{ \sqrt{ 1 + x^2 } \over 2 } + \displaystyle{ 1-x \over 3 } $

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

Now differentiate this equation, getting

$ T' = (1/2)(1/2) \big\{ 1 + x^2 \big\}^{-1/2} (2x) - 1/3 $

$ = \displaystyle{ x \over 2 \sqrt{ 1 + x^2 } } - \displaystyle{ 1 \over 3 } $

= 0 ,

so that

$ \displaystyle{ x \over 2 \sqrt{ 1 + x^2 } } = \displaystyle{ 1 \over 3 } $

and

$ 3x = 2 \sqrt{ 1 + x^2 } $ .

Square both sides of this equation, getting

9x2 = 4 (1 + x2) = 4 + 4x2 ,

so that

5x2 = 4 ,

x2 = 4/5 ,

or

$ x = \displaystyle{ \pm 2 \over \sqrt{5} } \approx \pm 0.89 $ .

But $ x \ne \displaystyle{ - 2 \over \sqrt{5} } $ since variable x measures a distance and $ 0 \le x \le 1 $ . See the adjoining sign chart for T' .

If

$ x= \displaystyle{ 2 \over \sqrt{5} } \approx 0.89 $ mi.

then

$ T \approx 0.71 $ hr.

is the shortest possible time of travel.

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SOLUTION 10 : Let variable x be the width and variable y the length of the rectangular portion of the window.

The semi-circular portion of the window has length

$ (1/2) \pi $ (radius) $^2 = (1/2) \pi (x/2)^2 = ( \pi /8) x^2 $ .

The perimeter (distance around outside only) of the window is given to be

$ 12 = x + 2y + ( \pi /8) x^2 $

so that

$ 2y = 12 - x - ( \pi /8) x^2 $

or

$ y = 6 - (1/2) x - ( \pi /16) x^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 = x y

$ = x ( 6 - (1/2) x - ( \pi /16) x^2 ) $

$ = 6x - (1/2) x^2 - ( \pi /16) x^3 $ .

Now differentiate this equation, getting

$ A' = 6 - x - (3 \pi /16) x^2 $

$ = (-3 \pi /16) x^2 - x + 6 $

= 0

for

$ x= \displaystyle{ -(-1) \pm \sqrt{ (-1)^2-4(-3 \pi/16)(6) } \over 2 (-3 \pi/16) } $ ,

i.e.,

$ x \approx 2.45 $ .

Since variable x measures distance, $ x \ge 0 $ . In addition, x is largest when y = 0 and the window is in the shape of a semi-circle. Thus, $ 0 \le x \le 4.40 $ (Why ?). See the adjoining sign chart for A' .

If

$ x \approx 2.45 $ ft. and y=3 ft. ,

then

$ A \approx 8.81 $ ft.2

is the largest possible area of the rectangle.

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SOLUTION 11 : Let variable x be the ADDITIONAL trees planted in the existing orchard. We wish to MAXIMIZE the total PRODUCTION of apples

P = (number of trees) (apple output per tree)

= ( 50 + x ) ( 800 - 10x )

= 40,000 + 300 x - 10 x2 .

Now differentiate this equation, getting

P' = 300 - 20 x

= 20 ( 15 - x )

= 0

for

x=15 .

See the adjoining sign chart for P' .

If

x = 15 additional trees ,

then

P = 42,250 apples

is the largest possible production of apples.

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SOLUTION 12 : Let variable x be the length of the base and variable y the height of the inscribed rectangle.

We wish to MAXIMIZE the total AREA of the rectangle

A = (length of base) (height) = xy .

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

A = x y

= x ( 8 - x3 )

= 8x - x4 .

Now differentiate this equation, getting

A' = 8 - 4 x3

= 4 ( 2 - x3 )

= 0 ,

so that

x3 = 2

and

$ x=2^{1/3} \approx 1.26 $ .

Note that $ 0 \le x \le 2 $ . See the adjoining sign chart for A' .

If

$ x=2^{1/3} \approx 1.26 $ and y = 6 ,

then

$ A = 6 ( 2^{1/3} ) \approx 7.60 $

is the largest possible area for the inscribed rectangle.

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SOLUTION 13 : Let variable r be the length of the base and variable h the height of the rectangle.

It is given that the perimeter of the rectangle is

12 = 2r + 2h

so that

2h = 12 - 2r

and

h = 6 - r .

We wish to MAXIMIZE the total VOLUME of the resulting CYLINDER

V = (area of base) (height) $ = ( \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 = \pi r^2 h $

$ = \pi r^2 ( 6 - r ) $

$ = \pi ( 6 r^2 - r^3 ) $ .

Now differentiate this equation, getting

$ V' = \pi ( 12 r - 3 r^2 ) $

$ = \pi (3 r )( 4 - r^2 ) $

$ = 3 \pi r ( 2 - r ) ( 2 + r ) $

= 0

for

r=0 or r=4 .

Since variable r measures distance and the perimeter of the rectangle is 12, $ 0 \le r \le 6 $ . See the adjoining sign chart for V' .

If

r = 4 ft. and h = 2 ft. ,

then

$ V= 16 \pi $ ft. $^3 \approx 50.27 $ ft.3

is the largest possible volume for the cylinder.

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SOLUTION 14 : Let variable $ \theta $ be the viewing angle and variable x the distance as denoted in the diagram. We seek to write angle $ \theta $ as a function of distance x . Introduce angle $ \alpha $ as in the diagram below.

It follows from basic trigonometry that

$ \tan \alpha = \displaystyle{ 10 \over x } $

so that

(Equation 1)

$ \alpha = \arctan \Big( \displaystyle{ 10 \over x } \Big) $ .

In a similar fashion

$ \tan ( \theta + \alpha ) = \displaystyle{ 30 \over x } $

so that

$ \theta + \alpha = \arctan \Big( \displaystyle{ 30 \over x } \Big) $ ,

or

(Equation 2)

$ \theta = \arctan \Big( \displaystyle{ 30 \over x } \Big) - \alpha $ .

Use $ \alpha $ from Equation 1 to substitute into Equation 2, getting

$ \theta = \arctan \Big( \displaystyle{ 30 \over x } \Big) - \arctan \Big( \displaystyle{ 10 \over x } \Big) $ .

We wish to MAXIMIZE angle THETA given in this equation. Differentiate this equation, getting

$ \theta' = \displaystyle{ 1 \over 1 + \Big( \displaystyle{ 30 \over x } \Big)^2...
...aystyle{ 10 \over x } \Big)^2 }
\Big\{ \displaystyle{ -10 \over x^2 } \Big\} $

$ = \displaystyle{ -30 \over x^2 + 30^2 } + \displaystyle{ 10 \over x^2 + 10^2 } $

= 0 ,

so that

$ \displaystyle{ 30 \over x^2 + 30^2 } = \displaystyle{ 10 \over x^2 + 10^2 } $ ,

30 x2 + 3000 = 10 x2 + 9000 ,

20 x2 = 6000 ,

x2 = 300 ,

for

$ x = \pm \sqrt{ 300 } = \pm 10 \sqrt{3} \approx \pm 17.32 $ .

But $ x \ne - 10 \sqrt{3} $ since variable x measures distance and $ x \ge 0 $ . If $ x = 10 \sqrt{3} $ , then

$ \theta = \arctan \Big( \displaystyle{ 30 \over 10 \sqrt{3} } \Big)
- \arctan \Big( \displaystyle{ 10 \over 10 \sqrt{3} } \Big) $

$ = \arctan \Big( \displaystyle{ 3 \over \sqrt{3} } \Big)
- \arctan \Big( \displaystyle{ 1 \over \sqrt{3} } \Big) $

$ = \arctan \Big( \displaystyle{ \sqrt{3} } \Big)
- \arctan \Big( \displaystyle{ 1 \over \sqrt{3} } \Big) $

(These are well-known values from basic trigonometry.)

$ = \displaystyle{ \pi \over 3 } - \displaystyle{ \pi \over 6 } $

$ = \displaystyle{ \pi \over 6 } $ radians

or

$ \theta = 30 $ degrees .

See the adjoining sign chart for $ \theta' $ .

If

$ x = 10 \sqrt{3} $ ft. $ \approx 17.32 $ ft.

then

$ \theta = 30 $ degrees $ = \displaystyle{ \pi \over 6 } $ radians

is the largest possible viewing angle.

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Duane Kouba
1998-06-17