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Elden [556K]
3 years ago
14

Lucy is a dress maker. She sews 4/7 of a dress in 3/4 hour. Lucy sees at a constant rate. At this rate, how many dresses does lu

cy see in one hour? (Include fractions of dresses if applicable.)
Mathematics
1 answer:
IgorLugansk [536]3 years ago
5 0

Answer:

Lucy sews \frac{16}{21} dresses in 1 hour.

Step-by-step explanation:

Given:

Lucy sews \frac{4}{7} of a dress in \frac{3}{4} hour.

She sews at constant rate.

We need to find the number of dresses she makes in 1 hour.

In \frac{3}{4} hour = \frac{4}{7} dress

So in 1 hour = Number of dresses in 1 hour

By using Unitary method we get;

Number of dresses in 1 hour= \frac{\frac{4}{7}}{\frac{3}{4}} = \frac{4}{7}\times \frac{4}{3}= \frac{16}{21}

Hence Lucy sews \frac{16}{21} dresses in 1 hour.

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Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. f(x, y) = 2x2y,
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Answer:

f(-2,-1) = -8 is the minimum value of f(x,y)

f(2,1) = 8 is the maximum value of f(x,y)

Step-by-step explanation:

f(x,y) = 2x²y under the constraint 2x² + 4y² = 12

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df/dx + λdg/dx = 0    and df/dy + λdg/dy = 0.

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xy + xλ = 0

x(y + λ) = 0

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2x² = 8λ²

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12λ² = 12

λ² = 1

λ = ±1

Substituting the value of  λ into x and y, we have

x = ±2λ = ±2(±1) = ±2

y = -λ = -(±1) = ±1

The minimum values of x and y are -2 and -1 respectively. Substituting these int f(x,y), we have

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So f(-2,-1) = -8 is the minimum value of f(x,y)

The maximum values of x and y are 2 and 1 respectively. Substituting these int f(x,y), we have

f(2,1) = 2(2)²(1) = 2 × 4 × 1 = 8

So f(2,1) = 8 is the maximum value of f(x,y)

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