* 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

*r*^{2} + *z*^{2} = 2^{2}

so that

*z*^{2} = 4 - *r*^{2}

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 - *r*^{2} ) = 9 *r*^{4} - 48 *r*^{2} + 64 ,

64 - 16 *r*^{2} = 9 *r*^{4} - 48 *r*^{2} + 64 ,

32 *r*^{2} - 9 *r*^{4} = 0 ,

*r*^{2} ( 32 - 9 *r*^{2} ) = 0 ,

so that

*r* = 0

or

32 - 9 *r*^{2} = 0 ,

*r*^{2} = 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.

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* 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

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
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 (*x*^{2}+3) from the numerator and simplify the expression.)

(Divide out a factor of (*x*^{2}+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) .

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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,

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 64*x* 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 2*x* 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.

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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* = *x*^{2} + ( *x*^{2} -6*x* +13 ) = 2*x*^{2} -6*x* +13 .

Now differentiate, getting

*S*' = 4*x* -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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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.

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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

*w*^{2} + (6-*x*)^{2} = *x*^{2} ,

so that

*w*^{2} = *x*^{2} - (*x*^{2} - 12*x* + 36) = 12*x* - 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} = 3*x* - 18 ,

-2 (*x*^{2} - 6*x* + 9) = 3*x* - 18 ,

-2 *x*^{2} + 12*x* -18 = 3*x* - 18 ,

-2 *x*^{2} + 9*x* = 0 ,

*x* ( -2*x* + 9 ) = 0 ,

so that (If *AB* = 0 , then *A* = 0 or *B* = 0 .)

*x* = 0

or

( -2*x* + 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.

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