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Advocard [28]
2 years ago
8

George has $49 which he decides to spend on x and y. commodity x costs $5 per unit and commodity y costs $11 per unit. he has th

e utility function u(x, y) = 3x 2 + 6y 2 and he can purchase fractional units of x and y. george will choose
Mathematics
1 answer:
FromTheMoon [43]2 years ago
7 0

We are given the equations:

<span>5 x + 11 y = 49                    --> eqtn 1</span>

<span>u = 3 x^2 + 6 y^2               --> eqtn 2</span>

 

Rewrite eqtn 1 explicit to y:

11 y = 49 – 5 x

<span>y = (49 – 5x) / 11               --> eqtn 3</span>

 

Substitute eqtn 3 to eqtn 2:

u = 3 x^2 + 6 [(49 – 5x) / 11]^2

u = 3 x^2 + 6 [(2401 – 490 x + 25 x^2) / 121]

u = 3 x^2 + 14406/121 – 2940x/121 + 150x^2/121

u = 4.24 x^2 – 24.3 x + 119.06

Derive then set du/dx = 0 to get the maxima:

du/dx = 8.48 x – 24.3 = 0

solving for x:

8.48 x = 24.3

x = 2.87

 

so y is:

y = (49 – 5x) / 11 = (49 – 5*2.87) / 11

y = 3.15

 

Answer:

<span>George will choose some of each commodity but more y than x.</span>

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Ashwin helps paint a square mural in his classroom. Then he helps paint a mural in the hallway whose length is 7 feet longer and
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Ashwin paints 2 murals:

  1. A square mural in his classroom
  2. A mural in the hallway with length 7 feet longer and width 3 feet shorter

Also, x indicates the side length of the mural in the classroom. Since the classroom mural is square in shape, the length and breadth of the mural would be x.

Part A: Expression to represents the two binomials to be multiplied to find the area of the hallway mural

Area of the hallway mural = Length of the hallway mural × Breadth of the hallway mural

So Length of the hallway mural = Length of the classroom mural + 7

⇒ Length of the hallway mural = x + 7

Breadth of the hallway mural = Breadth of the classroom mural - 3

⇒ Breadth of the hallway mural = x - 3

Hence, x + 7 and x - 3 need to be multiplied to determine the area of the mural

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Area of the hallway mural = Length of the hallway mural × Breadth of the hallway mural

⇒ Area of the hallway mural = (x + 7) × (x-3)

⇒ Area of the hallway mural = x² + 7x - 3x - 21

⇒ Area of the hallway mural = x² + 4x - 21

Part C: Area of the hallway mural if each side of the classroom mural is 8 foot long

Putting the value x = 8 in the answer of Part B

⇒ Area of the hallway mural = 8² + 4×8 - 21

⇒ Area of the hallway mural = 64 + 4×8 - 21

⇒ Area of the hallway mural = 64 + 32 - 21

⇒ Area of the hallway mural = 75 square feet


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Since, the slope of the line L is \frac{7}{6} , so the slope of the line which is perpendicular to the given line L is \frac{-6}{7} as the product of \frac{-6}{7} \times \frac{7}{6}=-1.

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