Solutions to Maximum/Minimum Problems

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 ,

r = 2 .

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

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)

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 ,

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

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' .

$x= \displaystyle{ 2 \over \sqrt{5} } \approx 0.89$ 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)(Circumference) = (1/2)(2(pi)(radius))= (1/2)(2(pi)(x/2))= (pi/2)x

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

12 = x + 2y + (pi/2)x

so that

2y = 12 - x - (pi/2)x

y = 6 - (1/2)x - (pi/4)x

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' .

$x \approx 2.45$ 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' .

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' .

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

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

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

(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

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