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3 years ago

By graphing the system of constraints, find the values o f x and y that minimize the objective function.

Mathematics

2 answers:

-BARSIC- [3]3 years ago

8 0

Answer:

Option (A)

Step-by-step explanation:

Here, x+ 2y =8

c=x+3y

x≥2

y≥0

By considering the options (A) and (B)

For option (A)

x=8 and y=0

x+2y=8

put values of x and y in equation

8+2×0=8, which satisfies the equation

now c= 8+3×0=8

Now, for option (B)

x=2 and y=3

x+2y=8

put values of x and y

2+2×3=8, which satisfies the equation.

now c= 2+3×3=11

As we have to find the minimum value of c.

hence we take c=8

Hence, option (A) is correct.

svet-max [94.6K]3 years ago

5 0

Graph the inequalities to find the vertices of the shaded region: (2, 3) and (8, 0).

Now, evaluate the the function C = x + 3y at those vertices to find the minimum value.

C = x + 3y at (2, 3) ⇒ C = (2) + 3(3) ⇒ C = 2 + 9 ⇒ C = 11

C = x + 3y at (8, 0) ⇒ C = (8) + 3(0) ⇒ C = 8 + 0 ⇒ C = 8

The minimum value occurs at (8, 0) with a minimum of C = 8

Answer: A

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Travka [436]

Answer:

D = L/k

Step-by-step explanation:

Since A represents the amount of litter present in grams per square meter as a function of time in years, the net rate of litter present is

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dA/(L - Ak) = dt

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∫-kdA/-k(L - Ak) = ∫dt

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1/k㏑(L - Ak) = t + C

㏑(L - Ak) = kt + kC

㏑(L - Ak) = kt + C' (C' = kC)

taking exponents of both sides, we have

$L - Ak = e^{kt + C'} \\L - Ak = e^{kt}e^{C'}\\L - Ak = C"e^{kt} (C" = e^{C'} )\\Ak = L - C"e^{kt}\\A = \frac{L}{k} - \frac{C"}{k} e^{kt}$

When t = 0, A(0) = 0 (since the forest floor is initially clear)

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