### 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

(area of base) (height) ,

so that

.

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)

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

.

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

.

Now differentiate this equation, getting

(Get a common denominator and combine fractions.)

= 0 ,

so that (If , then A=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

is the least possible cost of construction.

Click HERE to return to the list of problems.

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

.

We wish to MINIMIZE the total TIME elapsed

T = (swim time) + (walk time)

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

.

Now differentiate this equation, getting

= 0 ,

so that

and

.

Square both sides of this equation, getting

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

so that

5x2 = 4 ,

x2 = 4/5 ,

or

.

But since variable x measures a distance and . See the adjoining sign chart for T' .

If

mi.

then

hr.

is the shortest possible time of travel.

Click HERE to return to the list of problems.

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

(radius) .

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

so that

or

.

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

.

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, (Why ?). See the adjoining sign chart for A' .

If

ft. and y=3 ft. ,

then

ft.2

is the largest possible area of the rectangle.

Click HERE to return to the list of problems.

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.

Click HERE to return to the list of problems.

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

.

Note that . See the adjoining sign chart for A' .

If

and y = 6 ,

then

is the largest possible area for the inscribed rectangle.

Click HERE to return to the list of problems.

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) .

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

.

Now differentiate this equation, getting

= 0

for

r=0 or r=4 .

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

If

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

then

ft. ft.3

is the largest possible volume for the cylinder.

Click HERE to return to the list of problems.

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

It follows from basic trigonometry that

so that

(Equation 1)

.

In a similar fashion

so that

,

or

(Equation 2)

.

Use from Equation 1 to substitute into Equation 2, getting

.

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

= 0 ,

so that

,

30 x2 + 3000 = 10 x2 + 9000 ,

20 x2 = 6000 ,

x2 = 300 ,

for

.

But since variable x measures distance and . If , then

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

radians

or

degrees .

See the adjoining sign chart for .

If

ft. ft.

then

degrees radians

is the largest possible viewing angle.

Click HERE to return to the list of problems.

Duane Kouba
1998-06-17