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महाराष्ट्र राज्य शिक्षण मंडळएचएससी वाणिज्य (इंग्रजी माध्यम) इयत्ता १२ वी

Fill in the blank: A road of 108 m length is bent to form a rectangle. If area of the rectangle is maximum, then its dimensions are _______. - Mathematics and Statistics

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प्रश्न

Fill in the blank:

A road of 108 m length is bent to form a rectangle. If the area of the rectangle is maximum, then its dimensions are _______.

रिकाम्या जागा भरा

उत्तर

A road of 108 m length is bent to form a rectangle. If area of the rectangle is maximum, then its dimensions are x = 27, y = 27.

Explanation:

Let the length and breadth of a rectangle be x and y.

∴ Perimeter of rectangle = 2(x + y) = 108

∴ x + y = 54

∴ y = 54 - x      ....(i)

Let A = Area of rectangle = x × y

= x (54 - x) = 54x - x2

Differentiating w.r.t. we get

`"dA"/"dt" = 54 - 2"x"`

Consider, `"dA"/"dt" = 0` 

∴ 54 - 2x = 0

∴ x = 27

∴ y = 27          ....[from (i)]

x = 27, y = 27

shaalaa.com
Application of Derivatives to Economics
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 4: Applications of Derivatives - Miscellaneous Exercise 4 [पृष्ठ ११४]

APPEARS IN

बालभारती Mathematics and Statistics 1 (Commerce) [English] 12 Standard HSC Maharashtra State Board
पाठ 4 Applications of Derivatives
Miscellaneous Exercise 4 | Q 2.4 | पृष्ठ ११४

संबंधित प्रश्‍न

A manufacturing company produces x items at the total cost of Rs (180 + 4x). The demand function of this product is P = (240 − x). Find x for which profit is increasing.


Find the elasticity of demand, if the marginal revenue is 50 and price is Rs 75.


The manufacturing company produces x items at the total cost of ₹ 180 + 4x. The demand function for this product is P = (240 – x). Find x for which revenue is increasing


Find the price, if the marginal revenue is 28 and elasticity of demand is 3.


If the demand function is D = `((p + 6)/(p − 3))`, find the elasticity of demand at p = 4.


If the demand function is D = 50 – 3p – p2. Find the elasticity of demand at p = 5 comment on the result


For the demand function D = 100 – `"p"^2/2`. Find the elasticity of demand at p = 6 and comment on the results.


A manufacturing company produces x items at a total cost of ₹ 40 + 2x. Their price is given as p = 120 – x. Find the value of x for which profit is increasing.


A manufacturing company produces x items at a total cost of ₹ 40 + 2x. Their price is given as p = 120 – x. Find the value of x for which also find an elasticity of demand for price 80.


If the elasticity of demand η = 1, then demand is ______.


If the average revenue is 45 and elasticity of demand is 5, then marginal revenue is ______.


A manufacturing company produces x items at a total cost of ₹ 40 + 2x. Their price per item is given as p = 120 – x. Find the value of x for which elasticity of demand for price ₹ 80.

Solution: Total cost C = 40 + 2x and Price p = 120 – x

p = 120 – x

∴ x = 120 – p

Differentiating w.r.t. p,

`("d"x)/("dp")` = `square`

∴ Elasticity of demand is given by η = `- "P"/x*("d"x)/("dp")`

∴ η = `square`

When p = 80, then elasticity of demand η = `square`


If elasticity of demand η = 0 then demand is ______.


If f(x) = x3 – 3x2 + 3x – 100, x ∈ R then f"(x) is ______.


If 0 < η < 1 then the demand is ______.


In a factory, for production of Q articles, standing charges are ₹500, labour charges are ₹700 and processing charges are 50Q. The price of an article is 1700 - 3Q. Complete the following activity to find the values of Q for which the profit is increasing.

Solution: Let C be the cost of production of Q articles.

Then C = standing charges + labour charges + processing charges

∴ C = `square` 

Revenue R = P·Q = (1700 - 3Q)Q = 1700Q- 3Q2

Profit `pi = R - C = square`

 Differentiating w.r.t. Q, we get

`(dpi)/(dQ) = square`

If profit is increasing , then `(dpi)/(dQ) >0`

∴ `Q < square` 

Hence, profit is increasing for `Q < square` 


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