Optimization Two of the fastest romantic relationships ended with the following equations:L.R; = 4D? – 402D + 3D - 6, R2 = (1/3)D - 200D + 7. R stands for the relationships 1 and 2 while D stands forthe days it took before breakup. Which of the following is/are correct?A. Relationship 1 lasted longer than Relationship 2.B. Relationship 2 lasted longer than Relationship 1.c. Relationship 1 lasted for just almost 1 week.d. Relationship 2's duration is still shorter than the first-pay out period of salary given every half the month.3. In HS Physics, an object thrown vertically upward will always start to descend once it reaches its maximum height.Determine the time of ascent if the initial speed is v, and the acceleration due to gravity is % of that of Earth's.A. vo/gB. 0.5v./gC. 2v./9D. Insufficient Data4. The total surface area of an industrial cylindrical autoclave is to be limited to 20 m. If it is constructed to maximizethe volume, what should be the ratio of the height to that of the radius?A. 1В. 2C 3D. 10/3E. None of these5. Based on the concept of the previous number, if someone asks you for help in constructing a cylindrical tank witha top cover area of 100 cm?, accounting the minimum surface area and maximum volume, what would yourecommend for its lateral area?

Given data: Equation for Relations is, R1=4D2 - 402D + 3D3 -6R2=13D3 - 200D + 7 Formula used: For…

Answered: 1 Two of the fastest romantic…

Physics for Scientists and Engineers, Technology Update (No access codes included)

9th Edition

ISBN:9781305116399

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter1: Physics And Measurement

Section: Chapter Questions

Problem 1.29P

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Optimization

Two of the fastest romantic relationships ended with the following equations:
L.
R; = 4D? – 402D + 3D - 6, R2 = (1/3)D - 200D + 7. R stands for the relationships 1 and 2 while D stands for
the days it took before breakup. Which of the following is/are correct?
A. Relationship 1 lasted longer than Relationship 2.
B. Relationship 2 lasted longer than Relationship 1.
c. Relationship 1 lasted for just almost 1 week.
d. Relationship 2's duration is still shorter than the first-pay out period of salary given every half the month.
3. In HS Physics, an object thrown vertically upward will always start to descend once it reaches its maximum height.
Determine the time of ascent if the initial speed is v, and the acceleration due to gravity is % of that of Earth's.
A. vo/g
B. 0.5v./g
C. 2v./9
D. Insufficient Data
4. The total surface area of an industrial cylindrical autoclave is to be limited to 20 m. If it is constructed to maximize
the volume, what should be the ratio of the height to that of the radius?
A. 1
В. 2
C 3
D. 10/3
E. None of these
5. Based on the concept of the previous number, if someone asks you for help in constructing a cylindrical tank with
a top cover area of 100 cm?, accounting the minimum surface area and maximum volume, what would you
recommend for its lateral area?

Expert Solution

Step 1

Given data:

Equation for Relations is,

$R_{1} = 4 D^{2} - 402 D + 3 D^{3} - 6 R_{2} = (\frac{1}{3}) D^{3} - 200 D + 7$

Formula used:

For minimum or maximum value of R,

$\frac{d R}{d D} = 0$

Step 2

Calculation:

For R₁,

$R_{1} = 4 D^{2} - 402 D + 3 D^{3} - 6$

Differentiate with respect to D

$\frac{d R_{1}}{d D} = 0 \frac{d (4 D^{2} - 402 D + 3 D^{3} - 6)}{d D} = 0 8 D - 402 + 9 D^{2} = 0$

Compare above equation with standard quadratic equation

From above,

a=9, b= 8, c= -402

Therefore, roots of the above equation is,

$D = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} D = \frac{- 8 \pm \sqrt{{(8)}^{2} - 4 (9) (- 402)}}{2 (9)} D = \frac{- 8 \pm \sqrt{14536}}{18} D = \frac{- 8 \pm 120.56}{18} D = 6.253 O R D = - 7.142$

Therefore, putting value for R1,

for D= 6.253,

$R_{1} = 4 {(6.253)}^{2} - 402 (6.253) + 3 {(6.253)}^{3} - 6 R_{1} = 39.100 - 2513.706 + 733.125 - 6 R_{1} = - 1747.481$

Which is impossible (days never negative)

for D=-7.142 ,

$R_{1} = 4 {(- 7.142)}^{2} - 402 (- 7.142) + 3 {(- 7.142)}^{3} - 6 R_{1} = 39.100 - 2513.706 + 733.125 - 6 R_{1} = 1976.21$ ------------------(1)

Similarly for R₂,

$R_{2} = (\frac{1}{3}) D^{3} - 200 D + 7$

Differentiate with respect to D,

$\frac{d R_{2}}{d D} = \frac{d}{d D} ((\frac{1}{3}) D^{3} - 200 D + 7) = 0 3 D^{2} - 200 + 0 = 0 3 D^{2} = 200 D^{2} = 66.66 D = 8.164 O R D = - 8.164$

Therefore, putting value for R2,

for D= 8.164,

$R_{2} = (\frac{1}{3}) {(8.164)}^{3} - 200 (8.164) + 7 R_{2} = - 1444.35$

Which is not possible.

for D= -8.164,

$R_{2} = (\frac{1}{3}) {(- 8.164)}^{3} - 200 (- 8.164) + 7 R_{2} = 1458.42$ --------------------(2)

Form equation (1) and (2),

R₁ > R₂

Therefore, relation 1 is lasted longer than relation (2)

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