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serg [7]
3 years ago
12

A woman drives an SUV that gets ​13 mi/gal (mpg). Her husband drives a hybrid that gets 65 mpg. Every​ week, they travel the sam

e number of miles. They want to improve their combined mpg. They have two options on how they can improve it.
Option​ 1: They can tune the SUV and increase its mileage by 2 mpg and keep the hybrid as it is.


Option​ 2: They can buy a new hybrid that gets 80 mpg and keep the SUV as it is.

Which option will give them a better combined​ mpg?
Mathematics
1 answer:
Nady [450]3 years ago
5 0

Answer:

Option 1 will give a better overall mileage.

Step-by-step explanation:

The husband and wife travel the same number of miles each week.

Let's say each one travels x miles in 1 week.

Let e = fuel economy in mpg

Let v = amount of fuel in gallons

Let d = distance in miles

e = d/v

ve = d

v = d/e

Before modifying cars:

Wife:

d = x

e = 13

v = d/e = x/13

Husband:

d = x

e = 65

v = d/e = x/65

Combined fuel they both use in 1 week:

x/13 + x/65 = 5x/65 + x/65 = 6x/65 = 0.0923x

Option 1:

Wife changes to:

d = x

e = 15

v = d/e = x/15

Husband remains the same.

Combined fuel they both use in 1 week:

x/15 + x/65 = 13x/195 + 3x/195 = 16x/195 = 0.08205x

Option 2:

Husband changes to:

d = x

e = 80

v = d/e = x/80

Wife remains the same.

Combined fuel they both use in 1 week:

x/13 + x/80 = 80x/1040 + 13x/1040 = 93x/1040 = 0.08942x

The total fuel used in 1 week by both in terms of the miles, x, driven by both are:

Original case: 0.0923x

Option 1: 0.08205x

Option 2: 0.08942x

The lowest amount of fuel is 0.08205x in option 1.

Answer: Option 1 will give a better overall mileage.

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Step-by-step explanation:

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Hope this helps.

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Consider the equation below. (If an answer does not exist, enter DNE.) f(x) = x4 ln(x) (a) Find the interval on which f is incre
Ainat [17]

Answer: (a) Interval where f is increasing: (0.78,+∞);

Interval where f is decreasing: (0,0.78);

(b) Local minimum: (0.78, - 0.09)

(c) Inflection point: (0.56,-0.06)

Interval concave up: (0.56,+∞)

Interval concave down: (0,0.56)

Step-by-step explanation:

(a) To determine the interval where function f is increasing or decreasing, first derive the function:

f'(x) = \frac{d}{dx}[x^{4}ln(x)]

Using the product rule of derivative, which is: [u(x).v(x)]' = u'(x)v(x) + u(x).v'(x),

you have:

f'(x) = 4x^{3}ln(x) + x_{4}.\frac{1}{x}

f'(x) = 4x^{3}ln(x) + x^{3}

f'(x) = x^{3}[4ln(x) + 1]

Now, find the critical points: f'(x) = 0

x^{3}[4ln(x) + 1] = 0

x^{3} = 0

x = 0

and

4ln(x) + 1 = 0

ln(x) = \frac{-1}{4}

x = e^{\frac{-1}{4} }

x = 0.78

To determine the interval where f(x) is positive (increasing) or negative (decreasing), evaluate the function at each interval:

interval                 x-value                      f'(x)                       result

0<x<0.78                 0.5                 f'(0.5) = -0.22            decreasing

x>0.78                       1                         f'(1) = 1                  increasing

With the table, it can be concluded that in the interval (0,0.78) the function is decreasing while in the interval (0.78, +∞), f is increasing.

Note: As it is a natural logarithm function, there are no negative x-values.

(b) A extremum point (maximum or minimum) is found where f is defined and f' changes signs. In this case:

  • Between 0 and 0.78, the function decreases and at point and it is defined at point 0.78;
  • After 0.78, it increase (has a change of sign) and f is also defined;

Then, x=0.78 is a point of minimum and its y-value is:

f(x) = x^{4}ln(x)

f(0.78) = 0.78^{4}ln(0.78)

f(0.78) = - 0.092

The point of <u>minimum</u> is (0.78, - 0.092)

(c) To determine the inflection point (IP), calculate the second derivative of the function and solve for x:

f"(x) = \frac{d^{2}}{dx^{2}} [x^{3}[4ln(x) + 1]]

f"(x) = 3x^{2}[4ln(x) + 1] + 4x^{2}

f"(x) = x^{2}[12ln(x) + 7]

x^{2}[12ln(x) + 7] = 0

x^{2} = 0\\x = 0

and

12ln(x) + 7 = 0\\ln(x) = \frac{-7}{12} \\x = e^{\frac{-7}{12} }\\x = 0.56

Substituing x in the function:

f(x) = x^{4}ln(x)

f(0.56) = 0.56^{4} ln(0.56)

f(0.56) = - 0.06

The <u>inflection point</u> will be: (0.56, - 0.06)

In a function, the concave is down when f"(x) < 0 and up when f"(x) > 0, adn knowing that the critical points for that derivative are 0 and 0.56:

f"(x) =  x^{2}[12ln(x) + 7]

f"(0.1) = 0.1^{2}[12ln(0.1)+7]

f"(0.1) = - 0.21, i.e. <u>Concave</u> is <u>DOWN.</u>

f"(0.7) = 0.7^{2}[12ln(0.7)+7]

f"(0.7) = + 1.33, i.e. <u>Concave</u> is <u>UP.</u>

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Korolek [52]
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