CBSE 12th Standard Maths Subject Application of Derivatives Case Study Questions With Solution 2021
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CBSE 12th Standard Maths Application of Derivatives Case Study Questions With Solution 2021
12th Standard CBSE
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Reg.No. :
Maths
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Western music concert is organised every year in the stadium that can hold 36000 spectators. With ticket price of Rs. 10, the average attendance has been 24000. Some financial expert estimated that price of a ticket should be determined by the function \(p(x)=15-\frac{x}{3000}\), where x is the number of tickets sold.
Based on the above information, answer the following questions.
(i) The revenue, R as a function of xcan be represented as(a) \(15 x-\frac{x^{2}}{3000}\) (b) \(15-\frac{x^{2}}{3000}\) (c) \(15 x-\frac{1}{30000}\) (d) \(15 x-\frac{x}{3000}\) (ii) The range of x is
(a) [24000, 36000] (b) [0, 24000] (c) [0, 36000] (d) none of these (iii) The value of xfor which revenue is maximum, is
(a) 20000 (b) 21000 (c) 22500 (d) 25000 (iv) When the revenue is maximum, the price of the ticket is
(a) Rs. 5 (b) Rs. 5.5 (c) Rs. 7 (d) Rs. 7.5 (v) How any spectators should be present to maximize the revenue?
(a) 21500 (b) 21000 (c) 22000 (d) 22500 -
A tin can manufacturer designs a cylindrical tin can for a company making sanitizer and disinfector. The tin can is made to hold 3 litres of sanitizer or disinfector.
Based on the above in formation, answer the following questions.
(i) If r cm be the radius and h em be the height of the cylindrical tin can, then the surface area expressed as a function of r as(a) \(2 \pi r^{2}\) (b) \(\sqrt{\frac{500}{\pi}} \mathrm{cm}\) (c) \(\sqrt[3]{\frac{1500}{\pi}} \mathrm{cm}\) (d) \(2 \pi r^{2}+\frac{6000}{r}\) (ii) The radius that will minimize the cost of the material to manufacture the tin can is
(a) \(\sqrt[3]{\frac{600}{\pi}} \mathrm{cm}\) (b) \(\sqrt{\frac{500}{\pi}} \mathrm{cm}\) (c) \(\sqrt[3]{\frac{1500}{\pi}} \mathrm{cm}\) (d) \(\sqrt{\frac{1500}{\pi}} \mathrm{cm}\) (iii) The height thatt will minimize the cost of the material to manufacture the tin can is
(a) \(\sqrt[3]{\frac{1500}{\pi}} \mathrm{cm}\) (b) \(2 \sqrt[3]{\frac{1500}{\pi}} \mathrm{cm}\) (c) \(\sqrt{\frac{1500}{\pi}}\) (d) \(2 \sqrt{\frac{1500}{\pi}}\) (iv) If the cost of material used to manufacture the tin can is Rs.100/m2 and \(\sqrt[3]{\frac{1500}{\pi}} \approx 7.8\) then minimum cost is approximately
(a) Rs. 11.538 (b) Rs. 12 (c) Rs. 13 (d) Rs. 14 (v) To minimize the cost of the material used to manufacture the tin can, we need to minimize the
(a) volume (b) curved surface area (c) total surface area (d) surface area of the base -
Nitin wants to construct a rectangular plastic tank for his house that can hold 80 ft 3 of water. The top of the tank is open. The width of tank will be 5 ft but the length and heights are variables. Building the tank cost Rs.20 per sq. foot for the base and Rs. 10 per square foot for the side.
Based on the above information, answer the following questions.(i) In order to make a least expensive water tank, Nitin need to minimize its
(a) Volume (b) Base (c) Curved surface area (d) Cost (ii) Total cost of tank as a function of h can' be' represented as
(a) c(h) = 100 h - 320 - 1600lh (b) (h) = 100 h - 320 h - 720 h2 (c) c(h) = 100 + 220 h + 1600 h2 (d) \(c(h)=100 h+320+\frac{1600}{h}\) (iii) Range of h is
(a) (3,5) (b) \((0, \infty)\) (c) (0,8) (d) (0,3) (iv) Value of h at which c(h) is minimum, is
(a) 4 (b) 5 (c) 6 (d) 6.7 (v) The cost ofleast expensive tank is
(a) Rs. 1020 (b) Rs. 1100 (c) Rs. 1120 (d) Rs. 1220 -
Shreya got a rectangular parallelopiped shaped box and spherical ball inside it as return gift. Sides of the box are x, 2x, and x/3, while radius of the ball is r.
Based on the above information, answer the following questions.
(i) If S represents the sum of volume of parallelopiped and sphere, then S can be written as(a) \(\frac{4 x^{3}}{3}+\frac{2}{2} \pi r^{2}\) \(\frac{2 x^{2}}{3}+\frac{4}{3} \pi r^{2}\) \(\frac{2 x^{3}}{3}+\frac{4}{3} \pi r^{3}\) \(\frac{2}{3} x+\frac{4}{3} \pi r\) (ii) If sum of the surface areas of box and ball are given to be constant k2 , then x is equal to
(a) \(\sqrt{\frac{k^{2}-4 \pi r^{2}}{6}}\) (b) \(\sqrt{\frac{k^{2}-4 \pi r}{6}}\) (c) \(\sqrt{\frac{k^{2}-4 \pi}{6}}\) (d) none of these (iii) The radius of the ball, when S is minimum, is
(a) \(\sqrt{\frac{k^{2}}{54+\pi}}\) (b) \(\sqrt{\frac{k^{2}}{54+4 \pi}}\) (c) \(\sqrt{\frac{k^{2}}{64+3 \pi}}\) (d) \(\sqrt{\frac{k^{2}}{4 \pi+3}}\) (iv) Relation between length of the box and radius of the ball can be represented as
(a) x = 2r (b) \(x=\frac{r}{2}\) (c) \(x=\frac{r}{2}\) (d) \(\sqrt{\frac{k^{2}}{4 \pi+3}}\) (v) Minimum value of S is
(a) \(\frac{k^{2}}{2(3 \pi+54)^{2 / 3}}\) (b) \(\frac{k}{(3 \pi+54)^{3 / 2}}\) (c) \(\frac{k^{3}}{3(4 \pi+54)^{1 / 2}}\) (d) none of these -
Kyra has a rectangular painting canvas a toatl area of 24ft2 which include a border of 0.5ft on the left,right and a border 0.75 ft on the bottom,top inside it.
Based on the above information, answer the following questions.
(i) If Kyra wants to paint in the maximum area: then she needs to maximize(a) Area of outer rectangle (b) Area of inner rectangle (c) Area of top border (d) None of these (ii) If x is the length of the outer rectangle, then area of inner rectangle in terms of x is
(a) \((x+3)\left(\frac{24}{x}-2\right)\) (b) \((x-1)\left(\frac{24}{x}+1.5\right)\) (c) \((x-1)\left(\frac{24}{x}-1.5\right)\) (d) \((x-1)\left(\frac{24}{x}\right)\) (iii) Find the range of x.
(a) \((1, \infty)\) (b) (1, 16) (c) \((-\infty, 16)\) (d) (-1, 16) (iv) If area of inner rectangle is m~imum, then x is equal to
(a) 2ft (b) 3ft (c) 4ft (d) 5 ft (v) If area of inner rectangle is maximum, then length and breadth of this rectangle are respectively
(a) 3ft,4.Sft (b) 4.5ft,Sft (c) 1ft,2ft (d) 2ft,4ft
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