Q72.A box open from top is made from a rectangular sheet of dimension a × b by cutting squares each of side x from each of the four corners and folding up the flaps. If the volume of the box is maximum, then x is equal to: (1) a+b+√a2+b2−ab (2) a+b−√a2+b2−ab 6 12 (3) a+b−√a2+b2−ab (4) a+b−√a2+b2+ab 6 6
What This Question Tests
This problem requires setting up a volume function based on given dimensions, then using differentiation to find the value of x that maximizes the volume of the box.
Concepts Tested
Formulas Used
Volume of cuboid = length × width × height
d/dx (f(x)) = 0 for extremum
📚 NCERT Sections This Tests
9.23 — (A) At What Distance Should The Lens Be Held From The Card Sheet In
Physics Class 12 · Chapter 9
9.23 (a) At what distance should the lens be held from the card sheet in Exercise 9.22 in order to view the squares distinctly with the maximum possible magnifying power? (b) What is the magnification in this case? (c) Is the magnification equal to the magnifying power in this case? Explain.
9.18 — For Fixed Distance S Between Object And Screen, The Lens Equation
Physics Class 12 · Chapter 9
9.18 For fixed distance s between object and screen, the lens equation does not give a real solution for u or v if f is greater than s/4. Therefore, fmax = 0.75 m.
9.27 — (A) M = ( Fo/Fe) = 28
Physics Class 12 · Chapter 9
9.27 (a) m = ( fO/fe) = 28 f O f O (b) m = 1 + = 33.6 f e 25 349 Reprint 2025-26 Physics 9.28 (a) fO + fe = 145 cm (b) Angle subtended by the tower = (100/3000) = (1/30) rad. Angle subtended by the image produced by the objective h h = = f O 140 Equating the two, h = 4.7 cm. (c) Magnification (magnitude) of the eye-piece = 6. Height of the final image (magnitude) = 28 cm. 9.29 The image formed by the larger (concave) mirror acts as virtual object for the smaller (convex) mirror. Parallel rays coming from the object at infinity will focus at a distance of 110 mm from the larger mirror. The distance of virtual object for the smaller mirror = (110 –20) = 90 mm. The focal length of smaller mirror is 70 mm. Using the mirror formula, image is formed at 315 mm from the smaller mirror. 9.30 The reflected rays get deflected by twice the angle of rotation of the mirror. Therefore, d/1.5 = tan 7°. Hence d = 18.4 cm. 9.31 n = 1.33 CHAPTER 10 10.1 (a) Reflected light: (wavelength, frequency, speed same as incident light) l = 589 nm, n = 5.09 ´ 1014 Hz, c = 3.00 ´ 108 m s–1 (b) Refracted light: (frequency same as the incident frequency) n = 5.09 ´ 1014Hz v = (c/n) = 2.26 × 108 m s–1, l = (v/n) = 444 nm 10.2 (a) Spherical (b) Plane (c) Plane (a small area on the surface of a large sphere is nearly planar). 10.3 (a) 2.0 × 108 m s–1 (b) No. The refractive index, and hence the speed of light in a medium, depends on wavelength. [When no particular wavelength or colour of light is specified, we may take the given refractive index to refer to yellow colour.] Now we know violet colour deviates more than red in a glass prism, i.e. nv > nr. Therefore, the violet component of white light travels slower than the red component. 1.2 10 – 2 0.28 10 – 3 10.4 m = 600 nm 4 14. 10.5 K/4 10.6 (a) 1.17 mm (b) 1.56 mm 10.7 0.15° 350 10.8 tan–1(1.5) ~ 56.3o Reprint 2025-26 Answers
📋 Question Details
- Chapter
- Applications of Derivatives
- Topic
- Maximizing Volume
- Year
- 2021
- Shift
- 27 Aug Shift 2
- Q Number
- Q72
- Type
- MCQ
- NCERT Ref
- Class 12 Mathematics Ch 6: Application of Derivatives
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