Maxima and Minima Class 12 | Application of Derivatives | CBSE 2025 Maths PYQs Solved – Part 1

This video from Infinite X Solution focuses on Maxima and Minima problems from the Application of Derivatives chapter. This is Part 1 of the PYQ-based practice session for Class 12 CBSE Maths. 📌 What You’ll Learn: Conditions for Maxima and Minima Solving board-level PYQs Important tips and tricks for exam success 📘 Watch Part 2 for more advanced PYQs! #MaximaMinima #ApplicationOfDerivatives #Class12Maths #CBSE2025Maths #MathsBoardPrep #InfiniteXSolution 00:10 Introduction to maxima and minima 00:54 Introduction to Absolute maxima and Absolute minima 04:55 Introduction to local maxima and local minima 11:46 First derivative test to find local maxima and local minima 26:06 Point of inflection 33:26 Second derivative test 39:40 Absolute maximum and absolute minimum value of a function 49:47 Q1 Find the least value of the function f(x) = ax + b/x a,b,x are positive 56:21 Q2 An open tank with a square base and vertical sides is to be constructed from a metal sheet so as to hold a given quantity of water. Show that the cost of material will be least when depth of the tank is half of its width. If the cost is to be borne by nearby settled lower income families,for whom water will be provided. 01:08:19 Q3 The sum of the perimeters of a circle and square is k, where k is some constant. Prove that the sum of their areas is least, when the side of the square is double the radius of the circle. 01:19:22 Q4 Find the minimum value of (ax + by), where xy = c². 01:24:41 Q5 Find the dimensions of the rectangle of perimeter 36 cm which will sweep out a volume as large as possible, when revolved about one of its side. Also, find the maximum volume. 01:33:28 Q6 A tank with rectangular base and rectangular sides, open at the top is to be constructed, so that its depth is 2 m and volume is 8 m³. If building of tank cost Rs 70 per sq m for the base and Rs 45 per sq m for sides. What is the cost of least expensive tank? 01:42:28 Q7 Prove that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is 2R / √3 Also, find the maximum volume 01:51:40 Q8 Find the point on the curve y² = 4x, which is nearest to the point (2,- 8). 01:59:18 Q9 Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is 4r /3 Also, find the maximum volume in terms of volume of the sphere. Or Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is 4r/3 Also, show that the maximum volume of the cone is 8/27 of the volume of the sphere. 02:10:37 Q10 A window is of the form of a semi-circle with a rectangle on its diameter. The total perimeter of the window is 10 m. Find the dimensions of the window to admit maximum light through the whole opening. Or A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window to admit maximum light through the whole opening. 02:19:03 Q11 Show that the surface area of a closed cuboid with square base and given volume is minimum, when it is a cube. 02:25:30 Q12 AB is the diameter of a circle and C is any point on the circle. Show that the area of ABC is maximum, when it is an isosceles triangle. 02:34:30 Q13 If the sum of length of the hypotenuse and a side of a right angled triangle is given, show that the area of the triangle is maximum, when the angle between them is pi /3 02:41:09 Q14 A metal box with a square base and vertical sides is to contain 1024 cm³. The material for the top and bottom costs Rs 5 per cm² and the material for the sides costs Rs 2.50 per cm2. Find the least cost of the box. 02:47:04 Q15 Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is cos^-¹ 1/√3.