Only question 65,66,67 needs answer. Thank you follows:C(q) =10,000,000+500q+0.04q² &R(q) = 2,500q – 0.01q²%3Da. Using the marginal approach, determine the profit-maximizing level of output q'.b. What is the maximum profit?Optimum & Inflection Points (62-67) - Draw the graphs of the following functions and/orrelations and identify the locations of the optimum and the inflection points using the 1st and the2nd derivatives.62) y = f(x)= / 4x – 10 = (4x–10)½sRelation only-363) y = f(x)=1- xRelation and Function, both64) y = f(x)=(2x-7)*Relation and Function, both65) y = f(x)= /10-4x = (10-–4x)½Relation onlyx²66) y = f(x)=1- x2Relation and Function, both67) y = f(x)=(2x-7)°3.Relation and Function, both68) Cost Minimization - The cost function of purchasing, owning and maintaining inventoryannually for a firm's product is C = f(q)=(4,860/q)+15q +75,000 where C denotes annualinventory cost in dollars, and q equals the number of units of products ordered each time thefirm replenishes the inventory.%3D%Da. Determine the ordo

Name: Only question 65,66,67 needs answer. Thank you follows:C(q) =10,000,00
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Description: Only question 65,66,67 needs answer. Thank you follows:C(q) =10,000,000+500q+0.04q² &R(q) = 2,500q – 0.01q²%3Da. Using the marginal approach, determine the profit-maximizing level of output q'.b. What is the maximum profit?Optimum & Inflection Points

(Note: We are entitled to solve only one question at a time) Given: y=fx=10-4x=10-4x12 To Find: i.…

Answered: 65) y = f(x)= /10–4x = (10–4x)½ %3D

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Only question 65,66,67 needs answer. Thank you

follows:
C(q) =10,000,000+500q+0.04q² &
R(q) = 2,500q – 0.01q²
%3D
a. Using the marginal approach, determine the profit-maximizing level of output q'.
b. What is the maximum profit?
Optimum & Inflection Points (62-67) - Draw the graphs of the following functions and/or
relations and identify the locations of the optimum and the inflection points using the 1st and the
2nd derivatives.
62) y = f(x)= / 4x – 10 = (4x–10)½s
Relation only
-
3
63) y = f(x)=
1- x
Relation and Function, both
64) y = f(x)=(2x-7)*
Relation and Function, both
65) y = f(x)= /10-4x = (10-–4x)½
Relation only
x²
66) y = f(x)=
1- x
2
Relation and Function, both
67) y = f(x)=(2x-7)°
3.
Relation and Function, both
68) Cost Minimization - The cost function of purchasing, owning and maintaining inventory
annually for a firm's product is C = f(q)=(4,860/q)+15q +75,000 where C denotes annual
inventory cost in dollars, and q equals the number of units of products ordered each time the
firm replenishes the inventory.
%3D
%D
a. Determine the ordo

Expert Solution

Step 1

(Note: We are entitled to solve only one question at a time)

Given:

$y = f (x) = \sqrt{10 - 4 x} = {(10 - 4 x)}^{\frac{1}{2}}$

To Find:

$(i) . Location of Optimum Points. (i i) . Location of Inflection Points.$

Concept:

$(i) . Critical Point: x = c is called critical point of f (x), if f' (c) = 0 or f' (c) is undefined. (i i) . Inflection Point: Necessary Condition: x_{0} is a point of inflection of f (x), and f (x) has second derivative in some neighborhood of x_{0}, which is continuous at the point x_{0} itself then, f'' (x_{0}) = 0 . Sufficient Condition: If the function f (x) is continuous and differentiable at x_{0} and has second derivative f'' (x_{0}) in some deleted δ - neighbourhood of x_{0} and if f'' (x) changes sign through the point x_{0}, then x_{0} is called inflection point of f (x) .$

Step 2

Explanation:

$We have: y = f (x) = \sqrt{10 - 4 x} = {(10 - 4 x)}^{\frac{1}{2}} Domain of f (x) : 10 - 4 x \geq 0 \Rightarrow 10 \geq 4 x \Rightarrow x \leq \frac{5}{2} \Rightarrow Domain of f (x) = x \leq \frac{5}{2} or (- \infty, 2.5] . range of f (x) : f (x) \geq 0 or [0, \infty)$

Step 3

$(i) . Now, to find optimal point firstly we will find the inflection Points as f (x) = \sqrt{10 - 4 x} \Rightarrow f' (x) = \frac{- 4}{2 \sqrt{10 - 4 x}} = - \frac{2}{\sqrt{10 - 4 x}} f' (x) is not defined when \sqrt{10 - 4 x} = 0 i.e. 10 - 4 x = 0 \Rightarrow x = \frac{5}{2} hence, the critical point of f (x) is x = \frac{5}{2} = 2.5 Hence, we have two monotone intervals (- \infty, 2.5) and (2.5, \infty) but since, domain of f (x) is (- \infty, 2.5) hence, f (x) has only one monotone interval (- \infty, 2.5) . Now, we will check the sign of f' (x) in the interval in (- \infty, 2.5) let x \in (- \infty, 2.5) let x = 1 \Rightarrow f' (1) = - \frac{2}{\sqrt{10 - 4 (1)}} = - \frac{2}{\sqrt{6}} = - 0.81649 < 0 \Rightarrow f' (1) < 0 \Rightarrow f (x) is decreasing in (- \infty, 2.5) . hence, f (x) attains it's minima at x = 2.5 \Rightarrow f (2.5) = \sqrt{10 - 4 (2.5)} = \sqrt{10 - 10} = 0 \Rightarrow f (2.5) = 0 Hence, the function f (x) attains only minima at (2.5, 0) .$

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