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slava [35]
3 years ago
15

The demand function for a certain make of exercise bicycle sold exclusively through cable television is p=9−0.02x−−−−−−−−√(0≤x≤4

50) where p is the unit price in hundreds of dollars and x is the quantity demanded per week. Compute the elasticity of demand and determine the range of prices corresponding to inelastic, unitary, and elastic demand. Hint: Solve the equation E(p)=1.

Mathematics
1 answer:
nirvana33 [79]3 years ago
7 0

Answer:

Step-by-step explanation:

  • The concept of demand is applied, demand may be elastic, inelastic and unitary as the detailed steps and calculations is as shown in the attached file.
  • For elastic demand, P > 1
  • For inelastic demand, P < 1
  • For unitary, P = 1

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The graph above is a transformation of the function x2 Write an equation for the function graphed above
lutik1710 [3]

Answer: y=0,5x²-x-1,5.

Step-by-step explanation:

(-1;0)\ \ \ \ (3;0)\ \ \ \ (1;-2)   \ \ \ \  y=ax^2+bx+c\ ?\\\left\{\begin{array}{ccc}a*(-1)^2+b*(-1)+c=0\\a*3^2+b*3+c=0\\a*1^2+b*1+c=-2\end{array}\right \ \ \ \  \ \left\{\begin{array}{ccc}a-b+c=0\ \ (1)\\9a+3b+c=0\ (2)\\a+b+c=-2\ (3)\end{array}\right  \\

We summarize (1) and (3):

2a+2c=-2\ |:2\\a+c=-1\ \ \ \ \Rightarrow\\a+b+c=-2\\(a+s)+b=-2\\-1+b=-2\\b=-1.

Substitute b=-1 into (2):

9a+3*(-1)+c=0\\9a-3+c=0\\9a+c=3\ \ (5).\\

From (5) subtract (4):

8a=4\ |:8\\a=0,5.\ \ \ \  \Rightarrow\\

Substitute a=0,5 into (4):

0,5+c=-1\\c=-1,5.

Hense:

y=0,5x²-x-1,5.

5 0
2 years ago
-27 - x = -51<br> ................
Katarina [22]

Answer:

24

Step-by-step explanation:

-27 - x = -51

+27     +27

-x = -24

x = 24

6 0
3 years ago
A set of 12 windup toys costs $9. What is the cost per windup toy?
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Answer:

0.75

Step-by-step explanation:

9 ÷ 12 = 0.75

hope this helps...

4 0
3 years ago
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The diagonals of a trapezoid are perpendicular <br> Always <br> Sometimes <br> Never
iren [92.7K]
The correct answer is Sometimes
4 0
3 years ago
EXAMPLE 5 Find the maximum value of the function f(x, y, z) = x + 2y + 11z on the curve of intersection of the plane x − y + z =
Taya2010 [7]

Answer:

\displaystyle x= -\frac{10}{\sqrt{269}}\\\\\displaystyle y= \frac{13}{\sqrt{269}}\\\\\displaystyle z = \frac{23\sqrt{269}+269}{269}

<em>Maximum value of f=2.41</em>

Step-by-step explanation:

<u>Lagrange Multipliers</u>

It's a method to optimize (maximize or minimize) functions of more than one variable subject to equality restrictions.

Given a function of three variables f(x,y,z) and a restriction in the form of an equality g(x,y,z)=0, then we are interested in finding the values of x,y,z where both gradients are parallel, i.e.

\bigtriangledown  f=\lambda \bigtriangledown  g

for some scalar \lambda called the Lagrange multiplier.

For more than one restriction, say g(x,y,z)=0 and h(x,y,z)=0, the Lagrange condition is

\bigtriangledown  f=\lambda \bigtriangledown  g+\mu \bigtriangledown  h

The gradient of f is

\bigtriangledown  f=

Considering each variable as independent we have three equations right from the Lagrange condition, plus one for each restriction, to form a 5x5 system of equations in x,y,z,\lambda,\mu.

We have

f(x, y, z) = x + 2y + 11z\\g(x, y, z) = x - y + z -1=0\\h(x, y, z) = x^2 + y^2 -1= 0

Let's compute the partial derivatives

f_x=1\ ,f_y=2\ ,f_z=11\ \\g_x=1\ ,g_y=-1\ ,g_z=1\\h_x=2x\ ,h_y=2y\ ,h_z=0

The Lagrange condition leads to

1=\lambda (1)+\mu (2x)\\2=\lambda (-1)+\mu (2y)\\11=\lambda (1)+\mu (0)

Operating and simplifying

1=\lambda+2x\mu\\2=-\lambda +2y\mu \\\lambda=11

Replacing the value of \lambda in the two first equations, we get

1=11+2x\mu\\2=-11 +2y\mu

From the first equation

\displaystyle 2\mu=\frac{-10}{x}

Replacing into the second

\displaystyle 13=y\frac{-10}{x}

Or, equivalently

13x=-10y

Squaring

169x^2=100y^2

To solve, we use the restriction h

x^2 + y^2 = 1

Multiplying by 100

100x^2 + 100y^2 = 100

Replacing the above condition

100x^2 + 169x^2 = 100

Solving for x

\displaystyle x=\pm \frac{10}{\sqrt{269}}

We compute the values of y by solving

13x=-10y

\displaystyle y=-\frac{13x}{10}

For

\displaystyle x= \frac{10}{\sqrt{269}}

\displaystyle y= -\frac{13}{\sqrt{269}}

And for

\displaystyle x= -\frac{10}{\sqrt{269}}

\displaystyle y= \frac{13}{\sqrt{269}}

Finally, we get z using the other restriction

x - y + z = 1

Or:

z = 1-x+y

The first solution yields to

\displaystyle z = 1-\frac{10}{\sqrt{269}}-\frac{13}{\sqrt{269}}

\displaystyle z = \frac{-23\sqrt{269}+269}{269}

And the second solution gives us

\displaystyle z = 1+\frac{10}{\sqrt{269}}+\frac{13}{\sqrt{269}}

\displaystyle z = \frac{23\sqrt{269}+269}{269}

Complete first solution:

\displaystyle x= \frac{10}{\sqrt{269}}\\\\\displaystyle y= -\frac{13}{\sqrt{269}}\\\\\displaystyle z = \frac{-23\sqrt{269}+269}{269}

Replacing into f, we get

f(x,y,z)=-0.4

Complete second solution:

\displaystyle x= -\frac{10}{\sqrt{269}}\\\\\displaystyle y= \frac{13}{\sqrt{269}}\\\\\displaystyle z = \frac{23\sqrt{269}+269}{269}

Replacing into f, we get

f(x,y,z)=2.4

The second solution maximizes f to 2.4

5 0
3 years ago
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