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3 years ago

Linear programming subjected to constraints

Mathematics

1 answer:

anyanavicka [17]3 years ago

8 0

Answer:

The maximum value of C is 14

Step-by-step explanation:

we have the following constraints

$x\geq 0$ ----> constraint A

$y\geq 0$ ---> constraint B

$2x+2y\leq 10$ ---> constraint C

$3x+y\leq 9$ ---> constraint D

Determine the area of the feasible region using a graphing tool

see the attached figure

The vertices of the feasible region are

$(0,0),(0,5),(2,3),(3,0)$

To find out the maximum value of the objective function C, substitute the value of x and the value of y of each vertices in the objective function an then compare the results

we have

$C=4x+2y$

For (0,0) ----> $C=4(0)+2(0)=0$

For (0,5) ----> $C=4(0)+2(5)=10$

For (2,3) ----> $C=4(2)+2(3)=14$

For (3,0) ----> $C=4(3)+2(0)=12$

therefore

The maximum value of C is 14

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In the figure, ∆BAT ≅ ∆CAT. Which statement is not true by CPCTC? ∠CAB ≅ ∠SAB ∠CAS ≅ ∠BAS

Studentka2010 [4]

Answer:

∠CAB ≅ ∠SAB is not true

Step-by-step explanation:

CPCTC means id ABC = XYZ, than A=X, AB=XY, C=Z and so on. Basically the number numbers that come in the same order as the other one is equal.

So ∠CAS ≅ ∠BAS is true but ∠CAB ≅ ∠SAB is not.

Hope this helps. If you have any follow-up questions, feel free to ask.

Have a great day! :)

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3 years ago

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Answer:

Step-by-step explanation:

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6 0

3 years ago

∆ABC is a right angled at B and D is a point on BC if AD = 18cm BD = 9cm and CD =4cm Find AC

Gemiola [76]

In triangle, ABD,

AD²= AB²+BD²

AB² = AD²-BD²

AB² = 18²-9² = 324-81 = 243

AB = √243

In triangle, ABC,

AC² = AB²+BC²

AC² = (√243)²+(13)²

AC² = 243+169

AC = √412

AC = 20.29

3 0

3 years ago

Please solve the problem

jek_recluse [69]

Treat the matrices on the right side of each equation like you would a constant.

Let 2X + Y = A and 3X - 4Y = B.

Then you can eliminate Y by taking the sum

4A + B = 4 (2X + Y) + (3X - 4Y) = 11X

==> X = (4A + B)/11

Similarly, you can eliminate X by using

-3A + 2B = -3 (2X + Y) + 2 (3X - 4Y) = -11Y

==> Y = (3A - 2B)/11

It follows that

$X=\dfrac4{11}\begin{bmatrix}12&-3\\10&22\end{bmatrix}+\dfrac1{11}\begin{bmatrix}7&-10\\-7&11\end{bmatrix} \\\\ X=\dfrac1{11}\left(4\begin{bmatrix}12&-3\\10&22\end{bmatrix}+\begin{bmatrix}7&-10\\-7&11\end{bmatrix}\right) \\\\ X=\dfrac1{11}\left(\begin{bmatrix}48&-12\\40&88\end{bmatrix}+\begin{bmatrix}7&-10\\-7&11\end{bmatrix}\right) \\\\ X=\dfrac1{11}\begin{bmatrix}55&-22\\33&99\end{bmatrix} \\\\ X=\begin{bmatrix}5&-2\\3&9\end{bmatrix}$