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marshall27 [118]
3 years ago
7

Jacob is working two summer jobs, making $10 per hour washing cars and making $8 per hour landscaping. In a given week, he can w

ork no more than 19 total hours and must earn at least $170. If Jacob worked 4 hours landscaping, determine the minimum number of whole hours washing cars that he must work to meet his requirements. If there are no possible solutions, submit an empty answer.
Mathematics
1 answer:
Burka [1]3 years ago
6 0

Answer:

Step-by-step explanation:

Let x represent the number of hours that that he spent washing cars.

Let y represent the number of hours that that he spent landscaping.

In a week, he can work no more than 19 total hours. This means that

x + y ≤ 19- - - - - - - - - - - -1

Jacob is working two summer jobs, making $10 per hour washing cars and making $8 per hour landscaping. He must earn at least $170. This means that

10x + 8y ≥ 170 - - - - - - - - - -2

If Jacob worked 4 hours landscaping, it means that

x + 4 ≤ 19

x ≤ 19 - 4

x ≤ 15

Also,

10x + 8 × 4 ≥ 170

10x + 32 ≥ 170

10x ≥ 170 - 32

10x ≥ 138

10x ≥ 138/10

x ≥ 13.8

The minimum number of hours is 14

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The solution to the differential equation is

X(s)=\cfrac 1{6}  -\cfrac {1}{11}e^{t}+\cfrac {61}{66}e^{-2t}\cos(\sqrt 2t)+\cfrac {97}{66}\sqrt 2 e^{-2t}\sin(\sqrt 2t)

Step-by-step explanation:

Applying Laplace Transform will help us solve differential equations in Algebraic ways to find particular  solutions, thus after applying Laplace transform and evaluating at the initial conditions we need to solve and apply Inverse Laplace transform to find the final answer.

Applying Laplace Transform

We can start applying Laplace at the given ODE

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So we will get

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Applying initial conditions and solving for X(s).

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s^2 X(s)-s-1+4(sX(s)-1)+6X(s)=\cfrac 1s -\cfrac1{s-1}

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s^2 X(s)-s-1+4sX(s)-4+6X(s)=\cfrac 1s -\cfrac1{s-1}

s^2 X(s)-s-5+4sX(s)+6X(s)=\cfrac 1s -\cfrac1{s-1}

Moving all terms that do not have X(s) to the other side

s^2 X(s)+4sX(s)+6X(s)=\cfrac 1s -\cfrac1{s-1}+s+5

Factoring X(s) and moving the rest to the other side.

X(s)(s^2 +4s+6)=\cfrac 1s -\cfrac1{s-1}+s+5

X(s)=\cfrac 1{s(s^2 +4s+6)} -\cfrac1{(s-1)(s^2 +4s+6)}+\cfrac {s+5}{s^2 +4s+6}

Partial fraction decomposition method.

In order to apply Inverse Laplace Transform, we need to separate the fractions into the simplest form, so we can apply partial fraction decomposition to the first 2 fractions. For the first one we have

\cfrac 1{s(s^2 +4s+6)}=\cfrac As + \cfrac {Bs+C}{s^2+4s+6}

So if we multiply both sides by the entire denominator we get

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We can replace that on the previous equation and multiply all terms by 6

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We can simplify a bit

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And by comparing coefficients we can tell the values of B and C

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So the separated fraction will be

\cfrac 1{s(s^2 +4s+6)}=\cfrac 1{6s} +\cfrac {-s/6-4/6}{s^2+4s+6}

We can repeat the process for the second fraction.

\cfrac1{(s-1)(s^2 +4s+6)}=\cfrac A{s-1} + \cfrac {Bs+C}{s^2+4s+6}

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So the separated fraction will be

\cfrac1{(s-1)(s^2 +4s+6)}=\cfrac {1/11}{s-1} + \cfrac {-s/11-5/11}{s^2+4s+6}

So far replacing both expanded fractions on the solution

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We can combine the fractions with the same denominator

X(s)=\cfrac 1{6s}  -\cfrac {1/11}{s-1}+\cfrac {-s/6-4/6+s/11+5/11+s+5}{s^2 +4s+6}

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Completing the square

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Applying Inverse Laplace Transform.

Since all terms are ready we can apply Inverse Laplace transform directly to each term and we will get

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