\displaystyle \textbf{Question 1: } \text{At a toy factory, the production cost per day for } x \text{ number of toys is}
\displaystyle \text{given by } C(x)=0.0001x^{3}-0.001x^{2}-20x+3500. \text{ Which of these represents the}
\displaystyle \text{approximate additional cost to produce } 201\text{st toy?  }
\displaystyle \text{(a) } C(200) \quad \text{(b) } AC(201) \quad \text{(c) } MC(200) \quad \text{(d) } MC(201)
\displaystyle \text{Answer:}
\displaystyle \text{(d) We know that, marginal cost is interpreted as the approximate cost of one }
\displaystyle \text{additional unit of output.}
\displaystyle \therefore MC(201)\text{ will represent the approximate additional} \text{cost to produce }201\text{st toy.}
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\displaystyle \textbf{Question 2: } \text{The marginal revenue of a manufacturing facility is given by the function }
\displaystyle MR(x)=408-17x. \text{ The manufacturer should limit the production to } 24 \text{ units. Why?}
\displaystyle \text{Answer:}
\displaystyle \text{On substituting }x=24\text{ into }MR(x)=408-17x,
\displaystyle \text{we have, }MR(24)=408-17\times24=408-408=0
\displaystyle \text{If more than }24\text{ units will be produced, then the marginal revenue will be negative }
\displaystyle \text{and manufacturer will face loss. So, the manufacturer should limit the } \text{production to }24\text{ units.}
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\displaystyle \textbf{Question 3: } \text{Vishwanath runs a company that produces and sells a particular model of laptop.}
\displaystyle \text{Its revenue function is given by } R(x)=4x^{3}+\frac{1}{5}x^{2}+x, \text{ where } x \text{ is the number}
\displaystyle \text{of laptops demanded and } R(x) \text{ is in rupees. What is the ideal unit price (in rupees)}
\displaystyle \text{for Vishwanath to charge for each laptop, if there is a demand for } 100 \text{ laptops?}
\displaystyle \text{Answer:}
\displaystyle\text{On substituting }x=100\text{ into the revenue function}
\displaystyle R(x)=4x^{3}+\frac{1}{5}x^{2}+x,\text{ we get}
\displaystyle R(100)=4\times(100)^{3}+\frac{1}{5}(100)^{2}+100
\displaystyle =4000000+2000+100
\displaystyle =4002100
\displaystyle \text{So, ideal unit price}=\text{average revenue}
\displaystyle =\frac{4002100}{100}
\displaystyle =\text{Rs }40021
\displaystyle \text{Hence, Vishwanath should charge }40021\text{ rupees for }  \text{each laptop.}
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\displaystyle \textbf{Question 4: } \text{Which relation is correct for break-even point, where } R(x) \text{ is revenue}
\displaystyle \text{function and } C(x) \text{ is cost function?  }
\displaystyle \text{(a) } R(x)>C(x) \quad \text{(b) } R(x)<C(x) \quad \text{(c) } R(x)=C(x) \quad \text{(d) } R(x)>2C(x)
\displaystyle \text{Answer:}
\displaystyle \text{(c) We know that the break-even point at which total }  \text{revenue is equal to total cost}
\displaystyle R(x)=C(x)
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\displaystyle \textbf{Question 5: } \text{Average revenue of a commodity is given by } AR(x)=a+\frac{b}{x}. 
\displaystyle \text{Find the demand function, when marginal revenue is zero.  }
\displaystyle \text{Answer:}
\displaystyle \text{Given, }AR(x)=a+\frac{b}{x}\quad ...(i)
\displaystyle \Rightarrow \frac{dR}{dx}=a+\frac{b}{x}
\displaystyle \Rightarrow dR=\left(a+\frac{b}{x}\right)dx
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int dR=\int\left(a+\frac{b}{x}\right)dx
\displaystyle \therefore R=ax+b\log x+C\quad ...(ii)
\displaystyle \text{When }R=0\text{ and }x=0
\displaystyle 0=a\times0+b\log 0+C
\displaystyle \Rightarrow C=0
\displaystyle R=ax+b\log x\quad \text{[from Eq. (ii)]}
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\displaystyle \textbf{Question 6: } \text{If } R(x)=36x+3x^{2}+5, \text{ then find actual revenue from selling the}
\displaystyle 50\text{th item.  }
\displaystyle \text{Answer:}
\displaystyle \text{Given, }AR(x)=a+\frac{b}{x}\quad ...(i)
\displaystyle \Rightarrow \frac{dR}{dx}=a+\frac{b}{x}
\displaystyle \Rightarrow dR=\left(a+\frac{b}{x}\right)dx
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int dR=\int\left(a+\frac{b}{x}\right)dx
\displaystyle \therefore R=ax+b\log x+C\quad ...(ii)
\displaystyle \text{When }R=0\text{ and }x=0
\displaystyle 0=a\times0+b\log 0+C
\displaystyle \Rightarrow C=0
\displaystyle R=ax+b\log x\quad \text{[from Eq. (ii)]}
\displaystyle \text{Given, }R(x)=36x+3x^{2}+5
\displaystyle \text{Actual revenue received on selling 50th item}
\displaystyle =R(50)-R(49)
\displaystyle =(36\times50+3(50)^{2}+5)-(36\times49+3(49)^{2}+5)
\displaystyle =(1800+7500+5)-(1764+7203+5)
\displaystyle =9305-8972=333
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\displaystyle \textbf{Question 7: } \text{A company sells its product for Rs } 20 \text{ per unit. Fixed costs for the}
\displaystyle \text{company is Rs } 45000 \text{ and variable costs is estimated to run } 25\% \text{ of total revenue.} 
\displaystyle \text{If } x \text{ denotes}\text{ number of units produced, then what will be the total cost function?  }
\displaystyle \text{(a) } 45000+5x \quad \text{(b) } 15000+4x \quad \text{(c) } 45000+2x \quad \text{(d) } 4500+20x
\displaystyle \text{Answer:}
\displaystyle \text{(a) Given, selling price of product = Rs }20\text{ per unit}
\displaystyle \text{Fixed cost = Rs }45000
\displaystyle \text{and variable cost is estimated to run 25\% of total revenue}
\displaystyle \text{Revenue function }(R)=px=20x
\displaystyle \text{Cost function }(C)=\text{Fixed cost + variable cost}
\displaystyle \therefore C=45000+\frac{25}{100}R
\displaystyle =45000+\frac{1}{4}(20x)
\displaystyle =45000+5x
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\displaystyle \textbf{Question 8: } \text{The demand function for the certain commodity is given by }
\displaystyle p=4000-100x. \text{ What will be the total revenue from the sale of } 3 \text{ units?  }
\displaystyle \text{Answer:}
\displaystyle \text{(a) Given, }p=4000-100x
\displaystyle \text{Total revenue }=px=4000x-100x^{2}
\displaystyle \text{At }x=3,
\displaystyle \text{Total revenue}=4000(3)-100(3)^{2}
\displaystyle =12000-900
\displaystyle =11100
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\displaystyle \textbf{Question 9: } \text{A company sells } x \text{ packets of biscuits each day at Rs } 10 \text{ a packet.}
\displaystyle \text{The cost of manufacturing these packets is Rs } 5 \text{ per packet plus a fixed daily}
\displaystyle \text{overhead cost of Rs 700. What will be the profit function?  }
\displaystyle \text{(a) } 6x-400 \quad \text{(b) } 5x-700 \quad \text{(c) } 10x-500 \quad \text{(d) } 5x-10
\displaystyle \text{Answer:}
\displaystyle \text{(b) Given, a company sells }x\text{ packets of biscuits each }  \text{day at Rs }10\text{ per packet.}
\displaystyle \text{Manufacturing cost of one packet = Rs }5
\displaystyle \text{Total fixed cost = Rs }700
\displaystyle \therefore \text{Revenue function, }R(x)=10x\quad ...(i)
\displaystyle \text{Cost function }C(x)=\text{Variable cost + Fixed cost}
\displaystyle \therefore C(x)=700+5x\quad ...(ii)
\displaystyle \text{We know that profit function }P(x)=R(x)-C(x)
\displaystyle \text{On substituting the values of }R(x)\text{ and }C(x)\text{ from}
\displaystyle \text{Eqs. (i) and (ii) respectively, we have}
\displaystyle P(x)=10x-(700+5x)
\displaystyle =10x-700-5x
\displaystyle =5x-700
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\displaystyle \textbf{Question 10: } \text{The cost function of a firm is given by } C(x)=3x^{2}-2x+6.
\displaystyle  \text{ The average cost } \text{of the firm at } x=3 \text{ is  }
\displaystyle \text{(a) } 11 \quad \text{(b) } 17 \quad \text{(c) } 9 \quad \text{(d) } 27
\displaystyle \text{Answer:}
\displaystyle \text{(c) Given, cost function, }C(x)=3x^{2}-2x+6
\displaystyle \text{We know that, average cost }(AC)=\frac{C}{x}
\displaystyle \text{where }C\text{ is cost function and }x=\text{Number of commodities.}
\displaystyle \therefore AC=\frac{3x^{2}-2x+6}{x}\quad ...(i)
\displaystyle \text{On substituting }x=3\text{ into Eq. (i), we have}
\displaystyle AC=\frac{3(3)^{2}-2(3)+6}{3}
\displaystyle =\frac{27-6+6}{3}
\displaystyle =\frac{27}{3}=9
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\displaystyle \textbf{Question 11: } \text{The demand function of a monopolist is given by }
\displaystyle x=100-4p. \text{ The quantity }\text{at which } MR=0 \text{ will be  }
\displaystyle \text{(a) } 25 \quad \text{(b) } 10 \quad \text{(c) } 50 \quad \text{(d) } 30
\displaystyle \text{Answer:}
\displaystyle \text{(c) Given that, }x=100-4p
\displaystyle \text{So, }p=25-\frac{x}{4}
\displaystyle \text{Here, }R=px=\left(25-\frac{x}{4}\right)x=25x-\frac{x^{2}}{4}
\displaystyle \text{Now, }MR=\frac{dR}{dx}
\displaystyle \Rightarrow 0=\frac{d}{dx}\left(25x-\frac{x^{2}}{4}\right)
\displaystyle \Rightarrow 0=25-\frac{2x}{4}
\displaystyle \Rightarrow 0=25-\frac{x}{2}
\displaystyle \Rightarrow \frac{x}{2}=25\Rightarrow x=50
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\displaystyle \textbf{Question 12: } \text{Find the marginal cost function } (MC), \text{ if the cost function is } 
\displaystyle C(x)=\frac{x^{3}}{3}+5x^{2}-16x+2.   
\displaystyle \text{Answer:}
\displaystyle \text{Given that, }C(x)=\frac{x^{3}}{3}+5x^{2}-16x+2
\displaystyle \text{Now, }MC=\frac{d}{dx}\{C(x)\}
\displaystyle =\frac{d}{dx}\left(\frac{x^{3}}{3}+5x^{2}-16x+2\right)
\displaystyle =\frac{3x^{2}}{3}+10x-16
\displaystyle =x^{2}+10x-16
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\displaystyle \textbf{Question 13: } \text{If revenue function } R(x)=3x^{3}+8x-2, \text{ then find the average revenue}
\displaystyle \text{function.  }
\displaystyle \text{Answer:}
\displaystyle \text{Given that, }R(x)=3x^{3}+8x-2
\displaystyle \text{So, Average revenue function }(p)=\frac{R(x)}{x}
\displaystyle =\frac{3x^{3}+8x-2}{x}
\displaystyle =3x^{2}+8-\frac{2}{x}
\\

\displaystyle \textbf{Question 14: } \text{A bicycle spare parts manufacturing company decided to }
\displaystyle \text{upgrade its production unit by installing advanced machinery. As a result of this }
\displaystyle \text{investment, the cost of producing spare parts increased. The company used to sell } 1000
\displaystyle \text{ spare parts daily at Rs } 50 \text{ per unit. } \text{The increased daily cost of production } C(x) \text{ for }
\displaystyle x \text{ number of spare parts is given by }C(x)=\frac{99}{2}x+525.
\displaystyle \text{If the selling price remains the same, find how many more spare parts must the }
\displaystyle \text{company sell daily to ensure no loss? }
\displaystyle \text{Answer:}
\displaystyle \text{The increased daily cost of production for}
\displaystyle 1000\text{ spare parts}=C(1000)=\frac{99}{2}\times1000+525
\displaystyle =99\times500+525
\displaystyle =49500+525=\text{Rs }50025
\displaystyle \text{Now, the company must have to sell Rs }50025\text{ of spare } \text{parts more daily at Rs }50
\displaystyle \text{ per unit to ensure no loss.}
\displaystyle \therefore \text{The required numbers of spare parts to be sold}
\displaystyle =\frac{50025}{50}=1000.5
\displaystyle \text{Hence, at least }1001\text{ more spare parts must be sold by }  \text{the company to ensure no loss.}
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\displaystyle \textbf{Question 15: } \text{The government's department of Culture is organising an art and}
\displaystyle \text{craft exhibition cum sale to promote local artists and their work. They want to charge}
\displaystyle \text{a nominal amount as entry fee. Profit, in rupees, generated by the entry tickets can be }
\displaystyle \text{calculated using the function } P(x)=-8x^{3}+29400x-60000, \text{ where } x \text{ is the price}
\displaystyle \text{of each ticket. What is the ideal price for each ticket that the department should set to }
\displaystyle \text{maximise the profit?  }
\displaystyle \text{Answer:}
\displaystyle \text{Given, }P(x)=-8x^{3}+29400x-60000
\displaystyle \therefore \frac{d(P(x))}{dx}=-24x^{2}+29400
\displaystyle \text{For profit to be maximum, put }\frac{d(P(x))}{dx}=0
\displaystyle \Rightarrow -24x^{2}+29400=0
\displaystyle \Rightarrow 24x^{2}-29400=0
\displaystyle \Rightarrow x^{2}=\frac{29400}{24}
\displaystyle \Rightarrow x^{2}=1225\Rightarrow x=\pm35
\displaystyle \text{Now, }\frac{d^{2}(P(x))}{dx^{2}}=\frac{d}{dx}(-24x^{2}+29400)=-48x
\displaystyle \text{At }x=-35,\;\frac{d^{2}P}{dx^{2}}=(-48)\times(-35)=1680>0
\displaystyle \text{At }x=35,\;\frac{d^{2}P}{dx^{2}}=-48\times35=-1680<0
\displaystyle \text{Hence, at }x=35,\text{ the profit will be maximum.}
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\displaystyle \textbf{Question 16: } \text{The total revenue function } R(x)=x+2x^{3}-3.5x^{2}. \text{ Find the point where}
\displaystyle \text{marginal revenue curve cuts the coordinate axis.  }
\displaystyle \text{Answer:}
\displaystyle \text{Given, total revenue function,}
\displaystyle R(x)=x+2x^{3}-3.5x^{2}
\displaystyle \therefore MR=\frac{d}{dx}(R(x))=\frac{d}{dx}(x+2x^{3}-3.5x^{2})
\displaystyle \Rightarrow MR=1+6x^{2}-7x
\displaystyle \text{Put }MR=0
\displaystyle \Rightarrow 1+6x^{2}-7x=0
\displaystyle \Rightarrow 6x^{2}-7x+1=0
\displaystyle \Rightarrow 6x^{2}-6x-x+1=0
\displaystyle \Rightarrow 6x(x-1)-1(x-1)=0
\displaystyle \Rightarrow (6x-1)(x-1)=0
\displaystyle \Rightarrow x=\frac{1}{6}\text{ or }x=1
\displaystyle \text{Hence, marginal revenue curve cuts the coordinate axis at }x=1,\frac{1}{6}.
\\

\displaystyle \textbf{Question 17: } \text{For the revenue function } R(x)=\frac{bx}{a+x}, \text{ show that the }
\displaystyle \text{marginal revenue function is increasing for all } b<0 \text{ and } a>0.   
\displaystyle \text{Answer:}
\displaystyle \text{Given, revenue function,}
\displaystyle R(x)=\frac{bx}{a+x}
\displaystyle \therefore \text{Marginal revenue }(MR)=\frac{d}{dx}\left(\frac{bx}{a+x}\right)
\displaystyle =\frac{(a+x)\frac{d}{dx}(bx)-bx\frac{d}{dx}(a+x)}{(a+x)^{2}}
\displaystyle =\frac{(a+x)b-bx(1)}{(a+x)^{2}}
\displaystyle =\frac{ab}{(a+x)^{2}}
\displaystyle \Rightarrow MR=ab(a+x)^{-2}
\displaystyle \text{To show that MR is increasing, we show that}
\displaystyle \frac{d}{dx}(MR)=\frac{-2ab}{(a+x)^{3}}
\displaystyle \text{Since, }b<0\text{ and }a>0,\text{ so }ab\text{ is negative and }-2ab\text{ is positive.}
\displaystyle \text{Also, }(a+x)^{3}\text{ is positive since }a\text{ and }x\text{ are positive.}
\displaystyle \text{Thus, }\frac{-2ab}{(a+x)^{3}}>0,\text{ so MR is increasing.}
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\displaystyle \textbf{Question 18: } \text{The total cost function for a production is given by } 
\displaystyle C(x)=\frac{3}{4}x^{2}-7x+27. \text{Find the number of units produced for which } 
\displaystyle MC=AC. \text{ } (MC=\text{Marginal Cost and } AC=\text{Average Cost}) 
\displaystyle \text{Answer:}
\displaystyle \text{We have, }C(x)=\frac{3}{4}x^{2}-7x+27
\displaystyle \text{Now, marginal cost }=\frac{d}{dx}(C(x))
\displaystyle =\frac{d}{dx}\left(\frac{3}{4}x^{2}-7x+27\right)
\displaystyle =\frac{3}{2}x-7
\displaystyle \text{and average cost }=\frac{C(x)}{x}=\frac{\frac{3}{4}x^{2}-7x+27}{x}
\displaystyle =\frac{3}{4}x-7+\frac{27}{x}
\displaystyle \text{According to the given condition,}
\displaystyle \text{Marginal Cost = Average Cost}
\displaystyle \Rightarrow \frac{3}{2}x-7=\frac{3}{4}x-7+\frac{27}{x}
\displaystyle \Rightarrow \frac{3}{2}x-\frac{3}{4}x=\frac{27}{x}
\displaystyle \Rightarrow \frac{3}{4}x=\frac{27}{x}
\displaystyle \Rightarrow x^{2}=\frac{27\times4}{3}
\displaystyle \Rightarrow x^{2}=36
\displaystyle \Rightarrow x=\pm6
\displaystyle \text{Since }x\text{ cannot be negative, }x=6
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\displaystyle \textbf{Question 19: } \text{Find the cost of increasing from } 100 \text{ to } 200 \text{ units if the marginal cost }
\displaystyle \text{(in rupees per unit) is given by the function } MC=0.003x^{2}-0.001x+25.   
\displaystyle \text{Answer:}
\displaystyle \text{We have, }MC=0.003x^{2}-0.01x+25
\displaystyle C(x)=\int MC\,dx
\displaystyle \text{Now, cost of increasing from }100\text{ to }200\text{ units is}
\displaystyle \int_{100}^{200}(0.003x^{2}-0.01x+25)\,dx
\displaystyle =\left[\frac{0.003x^{3}}{3}-\frac{0.01x^{2}}{2}+25x\right]_{100}^{200}
\displaystyle =0.001(8000000-1000000)-\frac{0.01}{2}(40000-10000)+25(200-100)
\displaystyle =7000-150+2500
\displaystyle =9350
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\displaystyle \textbf{Question 20: } \text{The total variable cost of manufacturing } x \text{ units in a firm is } 
\displaystyle \sqrt{3x+\frac{x^{5}}{25}}. \text{ Show that average variable cost increases with output } x.   
\displaystyle \text{Answer:}
\displaystyle \text{We have, total variable cost}=3x+\frac{x^{5}}{25}
\displaystyle \text{Average variable cost}=\frac{TVC}{x}
\displaystyle \therefore AVC=\frac{3x+\frac{x^{5}}{25}}{x}
\displaystyle \Rightarrow AVC=3+\frac{x^{4}}{25}
\displaystyle \text{On differentiating w.r.t. }x,\text{ we get}
\displaystyle \frac{d}{dx}(AVC)=\frac{4x^{3}}{25}\geq0,\ \forall x
\displaystyle \text{Hence, }AVC\text{ increases with output }x.
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\displaystyle \textbf{Question 21: } \text{If the demand function is given by } x=\frac{600-p}{8}, \text{ where the price is}
\displaystyle \text{Rs. } p \text{ per unit} \text{and the manufacturer produces } x \text{ unit per week at the total cost of }
\displaystyle x^{2}+78x+2500, \text{ find the value of } x \text{ for which the profit is maximum.  }
\displaystyle \text{Answer:}
\displaystyle \text{Given, demand function }x=\frac{600-P}{8}\Rightarrow P=600-8x
\displaystyle \text{Revenue function }R(x)=Px=(600-8x)x=600x-8x^{2}
\displaystyle \text{Also, cost function }C(x)=x^{2}+78x+2500
\displaystyle \text{Now, profit function }P(x)=R(x)-C(x)
\displaystyle \Rightarrow P(x)=600x-8x^{2}-x^{2}-78x-2500
\displaystyle \Rightarrow P(x)=-9x^{2}+522x-2500
\displaystyle \text{On differentiating w.r.t. }x,\text{ we get}
\displaystyle P'(x)=-18x+522
\displaystyle \text{For maxima or minima put }P'(x)=0
\displaystyle \Rightarrow -18x+522=0
\displaystyle \Rightarrow x=\frac{522}{18}=29
\displaystyle \text{and }P''(x)=-18<0
\displaystyle \therefore \text{Profit is maximum when }x=29
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\displaystyle \textbf{Question 22: } \text{The fixed cost of new product is Rs } 35000 \text{ and the variable cost per unit}
\displaystyle \text{is Rs 500. If the demand function: } P=5000-100x, \text{ find the break-even value(s)?  }
\displaystyle \text{Answer:}
\displaystyle \text{We have, fixed cost = Rs }35000
\displaystyle \text{Variable cost per cent = Rs }500
\displaystyle \text{Total cost of }x\text{ units = Fixed cost + Total variable cost of }x\text{ unit}
\displaystyle \therefore TC=35000+500x
\displaystyle \text{Also, we have demand function, }p=5000-100x
\displaystyle \therefore \text{Revenue function }R(x)=px
\displaystyle R(x)=(5000-100x)x
\displaystyle =5000x-100x^{2}
\displaystyle \text{Profit function }P(x)=R(x)-TC
\displaystyle P(x)=5000x-100x^{2}-35000-500x
\displaystyle =-100x^{2}+4500x-35000
\displaystyle \text{At break-even value, }P(x)=0
\displaystyle \Rightarrow -100x^{2}+4500x-35000=0
\displaystyle \Rightarrow x^{2}-45x+350=0\quad [\text{divide by }-100]
\displaystyle \Rightarrow (x-35)(x-10)=0
\displaystyle \Rightarrow x=35,10
\displaystyle \text{Hence, the break-even values are }x=35\text{ and }x=10
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