[maxima minima problems][section 16]

(1.) A ball is thrown upward from ground level. After t

seconds, its height (in feet) above the ground is

48t - 16t².

(a.) What is its initial vertical velocity?

h(t) = 48t - 16t² here is the problem

v(t) = 48 - 32t take the derivative, that's velocity

v(0) = 48 - 32(0) replace t with 0

v(0) = 48 multiply combine like terms

result: 48 feet per second

(b.) After how many seconds will the ball reach its maximum

height?

48 - 32t = 0 set the derivative equal to 0

+32t +32t add 32t to each side
__________________
48 = 32t add
___ ___
32 32 divide each side by 32

1.5 = t divide and cancel

result: 1.5 seconds

(c.) How high will the ball go?

h(1.5) = 48(1.5) - 16(1.5)² replace t with 1.5

h(1.5) = 72 - 36 multiply

h(1.5) = 36 combine like terms

result: 36 feet

(2.) A farmer wishes to set aside one acre of land for corn

and wheat. To keep out the cows, he encloses a rectangular

field with a fence costing 50 cents per linear foot. In

addition, a fence running down the middle of the field is

needed; such a fence costs 1 dollar per foot. Given that

one acre = 43,560 ft², what dimensions should the field have

to minimize the total cost?

f = 2x + 3y this is the fence length equation

xy = 43,560 this is the area equation
__ ______
x x divide each side by x

y = 43,560/x cancel

C = .50(2x) + 1y + .50(2y) this is the cost function

C = x + 2y multiply combine like terms

C = x + 2(43,560/x) replace y with 43,560/x

C = x + (87,120/x) multiply

C' = 1 - (87,120/x²) take the derivative of C

1 - (87,120/x²) = 0 set the derivative equal to 0

x² - 87,120 = 0 multiply thru by x², cancel

+ 87,120 + 87,120 add this to each side
_____________________________

x² = 87,120 add

x = 296.16 take the square root of each side

y = 43,560/x use this equation to find y

y = 43,560/296.16 replace x with 296.16

y = 147.58 divide

results: x = 296.16 y = 147.58

(3.) A farmer wishes to divide 20 acres of land along a river

into 6 smaller plots by using one fence parallel to the

river and 7 fences perpendicular to it. Verify that the

total amount of fencing is minimized if the sum of the

lengths of the 7 cross fences equals the length of the one

parallel to the river.

f = x + 7y this is the fence length function

xy = 20 this is the area equation
__ ___
x x divide each side by x

y = 20/x cancel

f = x + 7(20/x) replace y with 20/x

f = x + (140/x) multiply

f' = 1 - (140/x²) take the derivative

1 - (140/x²) = 0 set the derivative equal to 0

x² - 140 = 0 multiply thru by x², cancel

+ 140 +140 add 140 to each side
____________________
x² = 140 add

x² = 4*35 factor
__
x = 2√35 take square roots
__ __
y = 20/(2√35) replace x with 2√35
__
y = 10/√35 divide

check: x = 7y
__ __
2√35 = 7(10/√35) make substitutions
__ __
2√35 = 70/√35 multiply
__
2(35) = 70 multiply each side by √35, cancel

70 = 70 multiply

(4.) V = (pi)r²h use the volume formula for cylinder

A = 2(pi)r² + 2(pi)rh use the surface area formula, too

(pi)r²h = 50 set the volume equal to 50
_______ ________
(pi)r² (pi)r² divide each side by this

h = 50/[(pi)r²] cancel

A = 2(pi)r² + 2(pi)r[50/((pi)r²)] make substitution

A = 2(pi)r² + (100/r) cancel and multiply

A' = 4(pi)r - (100/r²) take the derivative

4(pi)r - (100/r²) = 0 set the derivative equal to 0

4(pi)r³ - 100 = 0 multiply thru by r², cancel

+ 100 +100 add 100 to each side
__________________________
4(pi)r³ = 100 add
________ _____
4 4 divide each side by 4

(pi)r³ = 25 divide and cancel

r³ = 25/pi divide each side by pi, cancel

r = (25/pi)^1/3 raise each side to the 1/3 power

h = 50/[(pi)r²] use this equation to find h

h = (50/pi)r^-2 write r on top and use negative exp

h = (50/pi)(25/pi)^1/3 * ^-2 make substitution

h = (50/pi)(25/pi)^-2/3 multiply exponents

h = 2(25/pi)(25/pi)^-2/3 factor like this

h = 2(25/pi)^1/3 add exponents

results: r = (25/pi)^1/3 and h = 2(25/pi)^1/3

(5.) A wire 35 cm long is cut into two pieces. One piece is

bent in the shape of a square, and the other is bent in

the shape of a circle. How should the wire be cut to

maximize the total area enclosed by the pieces? How

should the wire be cut to minimize the total enclosed area?

(i.) x + y = 35 this is the wire length equation

y = 35 - x subtract x from each side

35 - x = 2(pi)r use the circumference formula
________ ________
2(pi) 2(pi) divide each side by 2(pi)

r = (35 - x)/(2pi) cancel

A = (x/4)² + (pi)r² this is the area function

A = (x/4)² + (pi)(35 - x)²/(4pi²) make substitutions

A = (x/4)² + [(35 - x)²/(4pi)] cancel pi's

A = (x/4)² + [(1225 - 70x + x²)/(4pi)] square the binomial

A = (1/16)x² + [(1225 - 70x + x²)/(4pi)] square 1/4

A = (1/8)x - (70/4pi) + (x/2pi) take the derivative

(1/8)x - (70/4pi) + (x/2pi) = 0 set the derivative equal to 0

(pi/8)x - (70/4) + (x/2) = 0 multiply thru by pi, cancel

[(pi/8) + (1/2)]x - (70/4) = 0 factoro like this

(pi + 4)x - 140 = 0 multiply thru by 8, cancel

+ 140 +140 add 140 to each side
___________________________
(pi + 4)x = 140 add
__________ _________
pi + 4 pi + 4 divide each side by this

x = (140)/(pi + 4) cancel

y = 35 - [140/(pi + 4)] replace x with this

y = [35(pi + 4)/(pi + 4)] - [140/(pi + 4)]

[multiply by (pi + 4)/(pi + 4)]

y = (35pi + 140 - 140)/(pi + 4) subtract fractions

y = [35pi/(pi + 4)] combine like terms

result: x = (140)/(pi + 4) and y = [35pi/(pi + 4)]

[this will result in a MINIMUM]

(ii.) For the maximum area, use the whole wire as the circle.

(6.) A Norman window is constructed from a rectangular sheet

of glass surmounted by a semicircular sheet of glass. The

light that enters through a window is proportional to the

area of the window. What are the dimensions of the

Norman window having a perimeter of 30 feet that admits the

most light?

2y + 4r + (pi)r = 30 this is the perimeter equation

2y + (4 + pi)r = 30 factor

y + (1/2)(4 + pi)r = 15 multiply thru by 1/2, cancel

- (1/2)(4 + pi)r - (1/2)(4 + pi)r subt this fr ea side
____________________________________________
y = 15 - (1/2)(4 + pi)r subtract

A = 2ry + (1/2)(pi)r² this is the area function

A= 2r[15 - (1/2)(4 + pi)r] + (1/2)(pi)r² make substitution for y

A = 30r - (4 + pi)r² + (1/2)(pi)r² multiply thru parentheses

A' = 30 - 2(4 + pi)r + (pi)r take the derivative

30 - 2(4 + pi)r + (pi)r = 0 set the derivative equal to 0

30 - [2(4 + pi) - pi]r = 0 factor

-30 + [2(4 + pi) - pi]r = 0 multiply thru by -1

+ 30 + 30 add 30 to each side
_______________________________
[2(4 + pi) - pi]r = 30 add

r = 30/[2(4 + pi) - pi] divide each side by this, cancel

r = 30/(8 + 2pi - pi) multiply thru parentheses

r = 30/(8 + pi) combine like terms

y = 15 - (1/2)(4 + pi)r use this equation to find y

y = 15 - (1/2)(4 + pi)[30/(8 + pi)] replace r with this

y = 15 - [15(4 + pi)/(8 + pi)] multiply 1/2 by 30

y = 15[(8 + pi)/(8 + pi)] - [15(4 + pi)/(8 + pi)]

[multiply by (8 + pi)/(8 + pi)]

y = (120 + 15pi - 60 - 15pi)/(8 + pi) multiply thru subtract

fractions

y = (60)/(8 + pi) combine like terms

results: r = 30/(8 + pi) and y = 60/(8 + pi)

(7.) Answer problem 5 if the two pieces are to be formed

into the shape of a circle and an equilateral triangle.

x + y = 35 this is the wire length equation

-x - x subtract x from each side
_____________________
y = 35 - x subtract
_
A = x(√3/4) + (pi)r² this will be the area

35 - x = 2(pi)r use the circumference formula for circle
______ _______
2pi 2pi divide each side by 2pi

(35 - x)/(2pi) = r cancel

A = x(√3/4) + (pi)(35 - x)²
________________ make substitution
4(pi)²
_
A = x(√3/4) + (35 - x)²
_____________ cancel
4pi
_
A' = (√3/4) - 2(35 - x)
___________ take the derivative
4pi

_
(√3/4) - [2(35 - x)/(4pi)] = 0 set the derivative equal to 0
_
(pi)√3 - 2(35 - x) = 0 multiply thru by 4pi, cancel
_
(pi)√3 - 70 + 2x = 0 multiply thru parentheses
_ _
2x = 70 - (pi)√3 add 70 and -(pi)√3 to each side
_
x = 35 - (1/2)(pi)√3 multiply thru by 1/2, cancel

y = 35 - x use this equation to find y
_
y = 35 - [35 - (1/2)(pi)√3] make substitution
_
y = (1/2)(pi)√3 multiply thru and combine like terms
_ _
results: x = 35 - (1/2)(pi)√3 , y = (1/2)(pi)√3

(8.) A woman is on a lake in a canoe 1 km from the closest

point P of a straight shore line; she wishes to get to

a point Q, 10 km along the shore from P. To do so, she

paddles to a point R between P and Q and then walks the

remaining distance to Q. She can paddle 3 km/hr and she

can walk 5 km/hr. How should she pick the point R so

that she gets to Q as quickly as possible?

f(x) = (1/3)(1 + x²)^1/2 + (1/5)(10 - x) here is the problem

f(x) = (1/3)(1 + x²)^1/2 + 2 - 0.2x multiply thru parentheses

f'(x) = (x/3)(1 + x²)^-1/2 - 0.2 take the derivative

(x/3)(1 + x²)^-1/2 - 0.2 = 0 set the derivative equal to 0

(5x)(1 + x²)^-1/2 - 3 = 0 multiply thru by 15, cancel

+3 +3 add 3 to each side
________________________
(5x)(1 + x²)^-1/2 = 3 add

5x = 3(1 + x²)^1/2 multiply each side by (1 + x²)^1/2

and cancel

25x² = 9(1 + x²) square each side

25x² = 9 + 9x² multiply thru parentheses

-9x² - 9x² subtract 9x² from each side
_____________________
16x² = 9 subtract

4x = 3 take the square root of each side
___ __
4 4 divide each side by 4

x = 3/4 cancel

result: choose R to be 3/4 km from P.