### 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

.

Thus the height of the inscribed cone is

.

We wish to MINIMIZE the total VOLUME of the CONE

.

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 using the product rule and the chain rule, getting

(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.)

,

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

.

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.

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 only. Then

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 ,

and hence

or .

Because measures an angle in a triangle, it is logical to assume that . Thus, . See the adjoining sign chart for A' .

If

then

A = 9/2

is the largest possible area for the triangle.

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

.

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 (x2+3) from the numerator and simplify the expression.)

(Divide out a factor of (x2+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' .

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) .

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,

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

(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

,

(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.

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 = 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.

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.

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

= 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.

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

.

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

.

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 .)

.

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

x = 0

or

,

,

-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 . See the adjoining sign chart for L' .

If

x = 9/2 in. and in. in. ,

then

in. in.

is the length of the shortest possible crease.