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

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

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

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

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 $${ A \over B } = 0$$ , 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,

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

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 $${ A \over B } = 0$$ , 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

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 $y \approx 16.67$ ft. ,

then

$L \approx 17.64$ ft.

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 : 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

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

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 $${ A \over B } = 0$$ , then A = 0 .)

23,400 t - 5400 = 0 ,

and

$t \approx 0.23$ .

See the adjoining sign chart for L' .

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 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)

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

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

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

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

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 $${ A \over B } = 0$$ , 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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