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2 years ago

Find the maximum value of P=3x+2y subject to the following constraints: Please help me!!!

Mathematics

1 answer:

UkoKoshka [18]2 years ago

6 0

Answer:

maximum value of P is 33

Step-by-step explanation:

Sketch the lines represented by the constraints

5x + y = 16

with intercepts ( $\frac{16}{5}$ , 0) and (0, 16)

2x + 3y = 22

with intercepts (11, 0) and (0, $\frac{22}{3}$ )

The solutions to both are below the lines.

Solve 5x + y = 16 and 2x + 3y = 22 simultaneously to find the point of intersection at (2, 6)

The coordinates of the vertices of the feasible region are

(0, 0), (0, 16), (2, 6), 11, 0)

Evaluate the objective function at each vertex

P = 3(0) + 2(0) = 0 + 0 = 0

P = 3(0) + 2(16) = 0 + 32 = 32

P = 3(2) + 2(6) = 6 + 12 = 18

P = 3(11) + 2(0) = 33 + 0 = 33

The maximum value of P is 33 when x = 11 and y = 0

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you own a pet supply store, and the total sales in your trading area equal 27,000. Your market share is 45%. What were your sale

Mazyrski [523]

Answer:

The sales is Rs. 12150.

Step-by-step explanation:

Given : You own a pet supply store, and the total sales in your trading area equal 27,000. Your market share is 45%.

To find : What were your sales ?

Solution :

Total sales = 27,000

Market share = 45%

Sales is given by,

Sales is 45% of total sale.

So, $S=27000\times \frac{45}{100}$

$S=12150$

Therefore, the sales is Rs. 12150.

6 0

2 years ago

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Given △DEF, which is not equal to cos(F)? sin(F). sin(D). tan(F). cos(D)

valentina_108 [34]

Answer: The correct option is tan(F).

Explanation:

In the given figure the triangle is a right angle triangle because $\angle E=90^{\circ}$ .

It is also given that $DE=EF=5$ and $DF=5\sqrt{2}$ .

Since it is an isosceles right angle triangle, therefore the value of perpendicular and base is same for both angles D and F, which is 5.

$\cos (F)=\frac{Base}{Hypotenuse} =\frac{5}{5\sqrt{2}} =\frac{1}{\sqrt{2}}$

The value of cos(F) is $\frac{1}{\sqrt{2}}$ .

$\sin (F)=\frac{Perpendicular}{Hypotenuse} =\frac{5}{5\sqrt{2}} =\frac{1}{\sqrt{2}}$

The value of sin(F) is $\frac{1}{\sqrt{2}}$ .

$\sin (D)=\frac{Perpendicular}{Hypotenuse} =\frac{5}{5\sqrt{2}} =\frac{1}{\sqrt{2}}$

The value of sin(D) is $\frac{1}{\sqrt{2}}$ .

$\tan (F)=\frac{Perpendicular}{Base}=\frac{5}{5} =1$

The value of tan(F) is 1. Which is not equal to $\frac{1}{\sqrt{2}}$ .

$\cos (D)=\frac{Base}{Hypotenuse} =\frac{5}{5\sqrt{2}} =\frac{1}{\sqrt{2}}$

The value of cos(D) is $\frac{1}{\sqrt{2}}$ .

Therefore, the value of tan(F) is not equal to the value of cos(F), so the correct option is third.

3 0

2 years ago

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Explain when the median of a data set is a better measure of center than the mean

vlada-n [284]

The dot plot or histogram will be skewed.

The mean is pulled up or down toward the tail.

The mean is affected more than the median.

Sample Response: When there is an outlier in the data set, the dot plot or histogram will be skewed. In a skewed representation, the mean is pulled up or down toward the tail of the data. Therefore, skewed data affects the mean more than the median.

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2 years ago

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jack puts 1/3 pound of birdseed into his feeder every time he fills it. how many can jack fill his bird feeder with 4 pounds of

Flura [38]

Answer: Jack can fill his feeder 12 times with 4 pounds of birdseed.

Step-by-step explanation:

You need to analize the information given in the exercise, You know that every time Jack fills the feeder, he put $\frac{1}{3}$ pounds into it.

Then, in order to solve this exercise, let "x" represents the number of times that Jack can fill his feeder with 4 pounds of birdseed.

Keeping on mind the data provided in the exercise, you can set up de following proportion:

$\frac{1}{\frac{1}{3}} = \frac{x}{4}$

Finally, you must solve for "x" in order to find its value.

You get that this is:

$3= \frac{x}{4}\\\\(3)(4)=x\\\\x=12$

Therefore, you can conclude that Jack can fill his feeder 12 times with 4 pounds of birdseed.

4 0

2 years ago

Solve the simultaneous equation 3x²-xy=0, 2y-5x=1

ratelena [41]

Answer:

(x, y) = (0, 1/2) or (1, 3)

Step-by-step explanation:

The first equation factors as ...

x(3x -y) = 0

This has solutions x=0 and y=3x.

x = 0

Using this in the second equation gives ...

2y -0 = 1

y = 1/2

(x, y) = (0, 1/2) is a solution

y = 3x

Using the expression for y in the second equation, we get ...

2(3x) -5x = 1

x = 1 . . . . . . . . . simplify

y = 3x = 3 . . . . using x=1 in the first equation

(x, y) = (1, 3) is a solution

_____

Interestingly, the (red line) graph of 3x^2 -xy = 0 produced by this graphing calculator has a "hole" at x=0, It says that point is (0, undefined). In a sense, y is undefined, in that it can be anything. A more appropriate graph would graph that equation as the two lines x=0 and y=3x.

4 0

2 years ago