* 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 .)

,

*r*^{3} = 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.

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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)

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

9*x*^{2} = 4 (1 + *x*^{2}) = 4 + 4*x*^{2} ,

so that

5*x*^{2} = 4 ,

*x*^{2} = 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.

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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

(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 (Use the quadratic formula.)

,

i.e.,

or .

But
since variable *x* measures distance and
. 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 ft. ,

then

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 - 10*x* )

= 40,000 + 300 *x* - 10 *x*^{2} .

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.

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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 - *x*^{3} )

= 8*x* - *x*^{4} .

Now differentiate this equation, getting

*A*' = 8 - 4 *x*^{3}

= 4 ( 2 - *x*^{3} )

= 0 ,

so that

*x*^{3} = 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.

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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 = 2*r* + 2*h*

so that

2*h* = 12 - 2*r*

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 , *r*=2 , or *r*=-2 .

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

If

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

then

ft.
ft.^{3}

is the largest possible volume for the cylinder.

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* 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 *x*^{2} + 3000 = 10 *x*^{2} + 9000 ,

20 *x*^{2} = 6000 ,

*x*^{2} = 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.

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