हिंदी

The Total Cost C for Producing X Units is Rs (X2 + 60x + 50) and the Price is Rs (180 - X) per Unit. for How Many Units the Profit is Maximum. - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

The total cost C for producing x units is Rs (x2 + 60x + 50) and the price is Rs (180 - x) per unit. For how many units the profit is maximum.

योग

उत्तर

Revenue R = x (180 - x)

R = 180x - x2

Profit π(x) = R - C

= (180x - x2) - (x+ 60x + 50)

π(x) = -2x2 + 120x - 50

`(dπ)/(dx)` = -4x + 120

Profit π is maximum

If  `(dπ)/(dx)` = 0 and `(d^2y)/(dx^2)` < 0

`(dπ)/(dx)` = 0

-4x + 120 = 0

x = 30

Now, `(d^2π)/(dx^2)` = -4

`(d^2π)/(dx^2)` = -4 < 0at x = 30

Profit is maximum at x = 30 units.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
2016-2017 (July) Set 1

APPEARS IN

वीडियो ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्न

Discuss the continuity of the function f, where f is defined by `f(x) = {(-2,"," if x <= -1),(2x, "," if -1 < x <= 1),(2, "," if x > 1):}`


A function f(x) is defined as,

\[f\left( x \right) = \begin{cases}\frac{x^2 - x - 6}{x - 3}; if & x \neq 3 \\ 5 ; if & x = 3\end{cases}\]  Show that f(x) is continuous that x = 3.

Show that

\[f\left( x \right)\] = \begin{cases}\frac{x - \left| x \right|}{2}, when & x \neq 0 \\ 2 , when & x = 0\end{cases}

is discontinuous at x = 0.

 

Discuss the continuity of the following functions at the indicated point(s): 

\[f\left( x \right) = \left\{ \begin{array}{l}\frac{2\left| x \right| + x^2}{x}, & x \neq 0 \\ 0 , & x = 0\end{array}at x = 0 \right.\]

Determine the value of the constant k so that the function

\[f\left( x \right) = \begin{cases}k x^2 , if & x \leq 2 \\ 3 , if & x > 2\end{cases}\text{is continuous at x} = 2 .\]


Determine the values of a, b, c for which the function f(x) = `{((sin(a + 1)x + sin x)/x, "for"   x < 0),(x, "for"  x = 0),((sqrt(x + bx^2) - sqrtx)/(bx^(3"/"2)), "for"  x > 0):}` is continuous at x = 0.


If the functions f(x), defined below is continuous at x = 0, find the value of k. \[f\left( x \right) = \begin{cases}\frac{1 - \cos 2x}{2 x^2}, & x < 0 \\ k , & x = 0 \\ \frac{x}{\left| x \right|} , & x > 0\end{cases}\] 

 


Find the points of discontinuity, if any, of the following functions:  \[f\left( x \right) = \begin{cases}\frac{x^4 + x^3 + 2 x^2}{\tan^{- 1} x}, & \text{ if } x \neq 0 \\ 10 , & \text{ if }  x = 0\end{cases}\]


Find all the points of discontinuity of f defined by f (x) = | x |− | x + 1 |.


Find all point of discontinuity of the function 

\[f\left( t \right) = \frac{1}{t^2 + t - 2}, \text{ where }  t = \frac{1}{x - 1}\]

The function  \[f\left( x \right) = \begin{cases}\frac{e^{1/x} - 1}{e^{1/x} + 1}, & x \neq 0 \\ 0 , & x = 0\end{cases}\]

 


The points of discontinuity of the function\[f\left( x \right) = \begin{cases}\frac{1}{5}\left( 2 x^2 + 3 \right) , & x \leq 1 \\ 6 - 5x , & 1 < x < 3 \\ x - 3 , & x \geq 3\end{cases}\text{ is } \left( are \right)\]  


Find whether the function is differentiable at x = 1 and x = 2 

\[f\left( x \right) = \begin{cases}x & x \leq 1 \\ \begin{array} 22 - x  \\ - 2 + 3x - x^2\end{array} & \begin{array}11 \leq x \leq 2 \\ x > 2\end{array}\end{cases}\]

Examine the continuity of f(x)=`x^2-x+9  "for"  x<=3`

=`4x+3  "for"  x>3,  "at"  x=3` 


Find the points of discontinuity , if any for the function : f(x) = `(x^2 - 9)/(sinx - 9)`


Discuss the continuity of the function f at x = 0, where
f(x) = `(5^x + 5^-x - 2)/(cos2x - cos6x),` for x ≠ 0
      = `1/8(log 5)^2,`  for x = 0


Show that the function f defined by f(x) = `{{:(x sin  1/x",", x ≠ 0),(0",", x = 0):}` is continuous at x = 0.


If f(x) = `(sqrt(2) cos x - 1)/(cot x - 1), x ≠ pi/4` find the value of `"f"(pi/4)`  so that f (x) becomes continuous at x = `pi/4`


`lim_("x" -> "x" //4) ("cos x - sin x")/("x"- "x" /4)`  is equal to ____________.


The value of k (k < 0) for which the function f defined as

f(x) = `{((1-cos"kx")/("x"sin"x")","  "x" ≠ 0),(1/2","  "x" = 0):}`

is continuous at x = 0 is:


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×