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3 years ago

You have a babysitting job that pays $6 an hour. You also have

Mathematics

1 answer:

Natalka [10]3 years ago

6 0

Answer:

One possibility is to work for (10) hours as a babysitter, and (10) hours as a cashier.

Step-by-step explanation:

An easy way to solve this problem is to set up a system to model the situation. Create one equation to model the money make, and the other to model the time spent. Let parameters (x) and (y) represent the time one spends at each job.

Since one cannot spend more than (20) hours a week working, set the first equation, for time, equal to (20),

x + y = 20

Now multiply each unit for the time by the money earned at each job, set this new equation equal to (150), the minimum amount of money one wishes to earn,

6(x) + 9(y) = 150

Thus the system is the following,

$\left \{ {{x+y=20} \atop {6x+9y=150}} \right.$

Now use the process of elimination. The process of elimination is when one multiplies one of the equations by a term such that when one adds the two equations, one of the variables cancels. One can solve for the other variable, and then backsolve for the first variable. Multiply the first equation by (-6) so that the variable (x) cancels.

$\left \{ {{-6x-6y=-120} \atop {6x+9y=150}} \right.$

Add the two equations,

$3y=30$

Use inverse operations to solve for (y),

$y=10$

Now substitute the value of (y) back into one of the original equations and solve for (x),

$x+y=20$

$x+10=20$

$x=10$

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a) $t = 1$ , b) $t = -2$ , c) Concave upward: $(-\infty, 1)$ . No intervals being concave downward.

Step-by-step explanation:

a) Tangent line is vertical if slope is undefined, whose points are associated with discontinuities of the function. Rational functions have discontinuities associated with the denominator:

$f(t) = \frac{t^{2}+4\cdot t +4}{t^{2}-2\cdot t + 1}$

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There is no avoidable discontinuities. The only point where tangent line is vertical is $t = 1$ .

b) Tangent line is horizontal if slope is equal to zero, whose point is associated to the points that makes function equal to zero. Rational functions have horizontal tangent lines if numerator is equal to zero and denominator is different to zero. Hence, the only point where tangent line is horizontal is $t = -2$ .

c) An interval is concave upward if exist an absolute minimum inside, which can be found by the help of the First and Second Derivative Tests.

$f'(t) = \frac{2\cdot (t+2)\cdot (t-1)^{2}-2\cdot (t-1)\cdot (t+2)^{2}}{(t-1)^{4}}$

$f'(t) = 2\cdot (t+2)\cdot (t-1)\cdot \left[\frac{t-1-t-2}{(t-1)^{4}}\right]$

$f'(t) = -\frac{6\cdot (t+2)\cdot (t-1)}{(t-1)^{4}}$

$f'(t) = -\frac{6\cdot (t+2)}{(t-1)^{3}}$

The only critical point is:

$-6\cdot (t+2) = 0$

$t = -2$ (which coincides with the result of point b)

$f''(t) = -6\cdot \left[\frac{(t-1)^{3}-3\cdot (t-1)^{2}\cdot (t+2)}{(t-1)^{6}}\right]$

$f''(t) = -6\cdot \left[\frac{1}{(t-1)^{3}}-\frac{3\cdot (t+2)}{(t-1)^{4}} \right]$

The value associated with the critical point is:

$f''(-2) = \frac{2}{9}$

Which means that critical point is an absolute minimum, and, consequently, the interval that is concave upward is $(-\infty, 1)$ . There is no absolute maximums and, therefore, there is no interval that is concave downward.

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