CALCULUS 1
Name: _____________________________
WORKSHEET 4.5-2
Date: ______________ Block: ________
1. A manufacturer wants to make an open box. She will make it from a 15 inch by 15 inch piece of material by
cutting equal squares from each corner and turning up the sides. Find the dimensions and the volume of the
largest box that can be made.
V = lwh
V’ = 12x2 – 120x + 225
x = 5/2 in
V = 250 in3
V = (15 – 2x)(15 – 2x)x
0 = 3(4x2 – 40x + 75)
15 – 2(5/2) = 10 in
V = 225x – 60x2 + 4x3
0 = 3(2x – 5)(2x – 15)
15 – 2(5/2) = 10 in
2. A manufacturer wants to make an open box. She will make it from a 3 foot by 5 foot piece of material
by cutting equal squares from each corner and turning up the sides. Find the dimensions and the volume
of the largest box that can be made.
V = lwh
V = (3 – 2x)(5 – 2x)x
V = 15x – 16x2 + 4x3
x  .61 ft
V’ = 12x2 – 32x + 15
0 = 12x2 – 32x + 15
V = 4.10 ft3
3 – 2(.61) = 1.78 ft
Use quadratic formula!
5 – 2(.61) = 3.78 ft
3. A manufacturer wants to make an open box having a square base and a surface area of 192 square
centimeters. Find the dimensions and the volume of the largest box that can be made.
SA = x2 + 4xh
V = lwh
V’ = 48 – ¾ x2
192 = x2 + 4xh
V = xxh
0 = 48 – ¾ x2
𝐡=
𝟏𝟗𝟐−𝐱 𝟐
𝟒𝐱
𝟏𝟗𝟐−𝐱 𝟐
V = x2(
𝟒𝐱
)
64 = x2
V = 48x – ¼ x3
x = 8 cm
h = 4 cm
V = 256 cm3
x=8
4. A manufacturer wants to make an open box having a square base and a volume of 864 cubic meters. Find the
dimensions and the surface area of the box that uses the least amount of material.
V = lwh
SA = x2 + 4xh
864 = xxh
SA = x2 + 4x ( 𝐱𝟐 )
𝐡=
𝟖𝟔𝟒
𝐱𝟐
SA’ = 2x – 3456x-2
𝟖𝟔𝟒
0 = 2x -
𝟑𝟒𝟓𝟔
𝐱𝟐
SA = x2 + 3456x-10 = 2x3 - 3456
x3 = 1728
x = 12 m
h=6m
SA = 432 m2
5. Find the point on the graph of y  x 2  4 that is nearest to 2,0 .
How far is the distance between the two points?
x1 = x
d = √𝟐𝐱 𝟐 − 𝟒𝐱 + 𝟖
y = √(𝟏)𝟐 + 𝟒 = √𝟓
y1 = √𝐱 𝟐 + 𝟒
d’ = 4x – 4
d = √(𝟏 − 𝟐)𝟐 + (√𝟓 − 𝟎)𝟐 = √𝟔
x2 = 2
0 = 4x – 4
or
y2 = 0
x=1
d = √𝟐(𝟏)𝟐 − 𝟒(𝟏) + 𝟖 = √𝟔
6. Find the point on the graph of x 2  4 y 2  8 that is nearest to  1,0 . *Need to solve equation for y…y1
How far is the distance between the two points?
𝟑
d = √𝟒 𝐱 𝟐 + 𝟐𝐱 + 𝟑
x1 = x
𝟏
𝟑
y1 = √− 𝟒 𝐱 𝟐 + 𝟐
d’ = 𝟐x + 2
x2 = -1
0 = 𝟐x + 2
y2 = 0
x=-𝟑
𝟒 𝟐
𝟏
𝟏𝟒
y = √− 𝟒 (− 𝟑) + 𝟐 = √ 𝟗 =
𝟐
𝟒
√𝟏𝟒
d = √(− 𝟑 + 𝟏) + (
𝟑
𝟑
√𝟏𝟒
𝟑
𝟐
𝟏𝟓
− 𝟎) = √ 𝟗 =
√𝟏𝟓
𝟑
or
𝟒
𝟑
𝟒 𝟐
𝟒
𝟏𝟓
d = √𝟒 (− 𝟑) + 𝟐 (− 𝟑) + 𝟑 = √ 𝟗 =
√𝟏𝟓
𝟑
7. An oil field has 60 wells each producing on the average 200 barrels of oil per day. If new wells are added,
the average yield will drop by 2 barrels per day for each new well added. What number of well will give the
maximum daily yield from this field?
wells
60
60+x
80
barrels per day
200
2002x
160
Daily yield
12000
(60+x)(2002x)
12800
y = 12000 + 80x – 2x2
x = 20 wells added
y’ = 80 – 4x
8. A shipping company can ship a cargo of 100 tons at a profit of $5.00 per ton. If the company waits, the
shipment can be increased by 20 tons per week but the profit would decrease by $0.25 per ton.
How long should the shipping company wait? How much profit will the company earn?
tons
100
100+20x
250
Cost per ton
5.00
5-.25x
3.125
profit
500
(100+20x)(5-.25x)
$781.25
f(x) = 500 + 75x – 5x2
x = 7.5 weeks
f’(x) = 75 – 10x
P = $781.25