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3 years ago

5x y>_10 x y<_6 x 4y>_12 x>_0 y>_0 minimum for c=10,000x + 20,000y

Mathematics

1 answer:

Serga [27]3 years ago

5 0

We are given inequalities : 5x+y=>10

x+y<=6

x+4y=>12

x=>0

y=>0 .

Let us graph it first and the find the coordinate of the vertices of feasible region.

Coordinates of the feasible region are (1,5) , (1.474, 2.632) and (4,2).

Now, we need to plug those cordinates (1,5) , (1.474, 2.632) and (4,2) in the given function c=10,000x + 20,000y.

Let us plug those points one by one.

For (1,5)

c=10,000x + 20,000y = 10,000(1)+ 20,000(5)= 10,000+ 100,000 = 110,000.

For (1.474, 2.632)

C = 10,000(1.474)+ 20,000(2.632) = 67,380.

For (4,2)

C = 10,000(4)+ 20,000(2) = 80,000.

We got the minimum value 67,380 for (1.474, 2.632) coordinate.

Therefore, minimum for given function C=10,000x + 20,000y is 67,380.

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• (39) was addressed in 24434477

• (40) and (41) were both addressed in 24434541

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$\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dy$

and I assume C has a positive orientation in both cases

(a) It looks like the region has the curves y = x and y = x ² as its boundary***, so that the interior of C is the set D given by

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• Compute the line integral directly by splitting up C into two component curves,

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Then

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*** Obviously this interpretation is incorrect if the solution is supposed to be 3/2, so make the appropriate adjustment when you work this out for yourself.

• Compute the same integral using Green's theorem:

$\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dy = \iint_D \frac{\partial(4y-6xy)}{\partial x} - \frac{\partial(3x^2-8y^2)}{\partial y}\,\mathrm dx\,\mathrm dy \\\\ = \int_0^1\int_{x^2}^x 10y\,\mathrm dy\,\mathrm dx = \boxed{\frac23}$

(b) C is the boundary of the region

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• Compute the line integral directly, splitting up C into 3 components,

C₁ : x = t and y = 0 with 0 ≤ t ≤ 1

C₂ : x = 1 - t and y = t with 0 ≤ t ≤ 1

C₃ : x = 0 and y = 1 - t with 0 ≤ t ≤ 1

Then

$\displaystyle \int_C = \int_{C_1} + \int_{C_2} + \int_{C_3} \\\\ = \int_0^1 3t^2\,\mathrm dt + \int_0^1 (11t^2+4t-3)\,\mathrm dt + \int_0^1(4t-4)\,\mathrm dt \\\\ = \int_0^1 (14t^2+8t-7)\,\mathrm dt = \boxed{\frac53}$

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$\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dx = \int_0^1\int_0^{1-x}10y\,\mathrm dy\,\mathrm dx = \boxed{\frac53}$

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