8 Most FAQ’s of Maxima and Minima Chapter in Inter 1st Year Maths-1B (TS/AP)
7 Marks
LAQ-1 : Find two positive integers whose sum is 16 and the sum of whose squares is minimum
1) Given condition and expression for y:
$$x + y = 16 \quad \text{(Given)}$$
$$y = 16 – x \quad \text{(1)}$$
2) Expression for function f(x) in terms of x:
$$f(x) = x^2 + y^2 = x^2 + (16 – x)^2 \quad \text{(2)}$$
3) Differentiation of f(x) with respect to x:
$$f'(x) = 2x + 2(16 – x)(-1) = 2x – 32 + 2x$$
$$= 4x – 32 = 4(x – 8) \quad \text{(3)}$$
4) Condition for maxima or minima using derivative:
$$4(x – 8) = 0$$
$$x = 8$$
5) Substituting to find y using the value of x:
$$y = 16 – 8 = 8 \quad \text{from (1)}$$
6) Second derivative to determine the nature of critical point:
$$f”(x) = 4 \quad \text{(4)}$$
7) Evaluating second derivative at x=8:
$$\text{At } x = 8, f”(8) = 4 > 0$$
$$f(x) \text{ is minimum when } x = 8 \text{ and } y = 8$$
$$x = 8, y = 8$$
LAQ-2 : Find two positive integers whose sum is 15 and the sum of whose squares is minimum
1) Given condition and expression for y:
$$x + y = 15 \quad \text{(Given)}$$
$$y = 15 – x \quad \text{(1)}$$
2) Expression for function f(x) in terms of x:
$$f(x) = x^2 + y^2 = x^2 + (15 – x)^2 \quad \text{(2)}$$
3) Differentiation of f(x) with respect to x:
$$f'(x) = 2x + 2(15 – x)(-1) = 2x – 30 + 2x$$
$$= 4x – 30 = 2(2x – 15) \quad \text{(3)}$$
4) Condition for maxima or minima using derivative:
$$2(2x – 15) = 0$$
$$x = \frac{15}{2}$$
5) Substituting to find y using the value of x:
$$y = 15 – \frac{15}{2} = \frac{15}{2} \quad \text{from (1)}$$
$$f”(x) = 4 \quad \text{(4)}$$
6) Evaluating second derivative at x=15/2:
$$\text{At } x = \frac{15}{2}, f”\left(\frac{15}{2}\right) = 4 > 0$$
7) Conclusion about the nature of the critical point:
$$f(x) \text{ is minimum when } x = \frac{15}{2} \text{ and } y = \frac{15}{2}$$
$$x = \frac{15}{2}, y = \frac{15}{2}$$
LAQ-3 : Find two positive integers x and y such that x + y = 60 and xy3 is maximum
1) Given condition and expression for x:
$$x + y = 60 \quad \text{(Given)}$$
$$x = 60 – y \quad \text{(1)}$$
2) Expression for functionf(y) in terms of y:
$$\text{From (1) } f(y) = (60 – y)y^3$$
$$f(y) = 60y^3 – y^4 \quad \text{(2)}$$
3) Differentiation of f(y) with respect to y:
$$f'(y) = 180y^2 – 4y^3 \quad \text{(3)}$$
4) Setting the first derivative equal to zero to find critical points:
$$180y^2 – 4y^3 = 0$$
$$4y^2(45 – y) = 0$$
$$y = 0 \quad \text{or} \quad y = 45$$
$$\text{Also from (1) } x = 60 – y = 60 – 45 = 15$$
5) Second derivative to determine the nature of critical points:
$$f”(y) = 12y(30 – y) \quad \text{(4)}$$
6) Evaluating the second derivative at y=45:
$$f”(45) = 12(45)(30 – 45) = 12(45)(-15) = -12(45)(15)$$
$$\text{Thus, } f”(45) < 0$$
7) Conclusion about the nature of the critical point:
$$f(y) \text{has a maximum value at } y = 45 \text{ and } x = 15$$
Required integers are:
$$x = 15, y = 45$$
LAQ-4 : Find the maximum area of the rectangle that can be formed with fixed perimeter 20
1) Given perimeter condition and expression for y:
$$2(x + y) = 20 \quad \text{(Given perimeter)}$$
$$x + y = 10 \quad \text{(simplified)}$$
$$y = 10 – x \quad \text{(1)}$$
2) Expression for the area A of the rectangle in terms of x:
$$\text{Area } A = xy$$
$$\text{From (1) } A(x) = x(10 – x)$$
$$A(x) = 10x – x^2 \quad \text{(2)}$$
3) Differentiation of A(x) with respect to x to find the critical points:
$$A'(x) = 10 – 2x \quad \text{(3)}$$
4) Setting the first derivative equal to zero to find maximum or minimum:
$$10 – 2x = 0$$
$$x = 5$$
$$y = 10 – x = 10 – 5 = 5 \quad \text{from (1)}$$
5) Second derivative to determine the nature of the critical point:
$$A”(x) = -2 \quad \text{(4)}$$
6) Evaluating the second derivative at x=5:
$$\text{At } x = 5, A”(5) < 0$$
7) Conclusion about the nature of the critical point:
$$\text{Area is maximum at } x = 5 \text{ and } y = 5$$
$$A = xy = 5 \times 5 = 25 \text{ sq.units}$$
LAQ-5 : If the curved surface of right circular cylinder inscribed in a sphere of radius r is maximum, show that the height of the cylinder is √2r
$$\text{For the sphere given radius } = r$$
$$\text{For the cylinder we take height } = h, \text{ base radius } = R$$
$$AB = h, \quad BC = 2R, \quad AC = 2r$$
$$\text{Applying Pythagoras theorem on } \triangle ABC$$
$$h^2 + (2R)^2 = (2r)^2$$
$$h^2 + 4R^2 = 4r^2$$
$$h^2 = 4(r^2 – R^2) \quad \text{(1)}$$
$$\text{Curved surface area of the cylinder } A = 2\pi Rh$$
$$\text{On squaring and taking } A^2 = f(R) \text{ we get}$$
$$f(R) = 4\pi^2 R^2 h^2 = 4\pi^2 R^2[4(r^2-R^2)] = 16\pi^2 R^2(r^2-R^2)$$
$$f(R) = 16\pi^2(r^2 R^2 – R^4) \quad \text{(2)}$$
$$f'(R) = 16\pi^2[2Rr^2 – 4R^3] \quad \text{(3)}$$
$$16\pi^2[2Rr^2 – 4R^3] = 0$$
$$2R(r^2 – 2R^2) = 0$$
$$r^2 – 2R^2 = 0$$
$$R^2 = \frac{r^2}{2}$$
$$h^2 = 4(r^2 – R^2) = 4\left(r^2 – \frac{r^2}{2}\right) = 2r^2$$
$$h = \sqrt{2}r$$
$$f”(R) = 16\pi^2[2r^2 – 12R^2] \quad \text{(4)}$$
$$f”\left(\frac{r^2}{2}\right) = 16\pi^2\left(2r^2 – 12 \frac{r^2}{2}\right) = -64\pi^2r^2$$
$$\text{Thus } f”\left(\frac{r^2}{2}\right) < 0$$
$$\text{The curved surface area has a maximum value when } R = \sqrt{\frac{r^2}{2}} \text{ and } h = \sqrt{2}r.$$
LAQ-6 : From a rectangular sheet of dimensions 30cm x 80cm, four equal squares of sides x cm are removed at the corners, and the sides are then turned up so as to form an open rectangular box. What is the value of x, so that the volume of the box is the greatest?
$$\text{height } h = x$$
$$\text{length } l = 80 – 2x$$
$$\text{breadth } b = 30 – 2x \quad \text{(1)}$$
$$\text{Volume } V = l \times b \times h = (80 – 2x)(30 – 2x)(x)$$
$$= 2(40 – x)^2(15 – x)(x)$$
$$= 4(40 – x)(15 – x)(x) = 4(600 – 40x – 15x + x^2)x$$
$$= 4(600 – 55x + x^2)x = 4(x^3 – 55x^2 + 600x)$$
$$V(x) = 4(x^3 – 55x^2 + 600x) \quad \text{(2)}$$
$$V'(x) = 4(3x^2 – 110x + 600) \quad \text{(3)}$$
$$4(3x^2 – 110x + 600) = 0$$
$$3x^2 – 110x + 600 = 0$$
$$3x(x – 30) – 20(x – 30) = 0$$
$$(3x – 20)(x – 30) = 0$$
$$3x = 20 \quad \text{or} \quad x = 30$$
$$x = \frac{20}{3} \quad \text{or} \quad x = 30$$
$$V”(x) = 4(6x – 110) \quad \text{(4)}$$
$$V”\left(\frac{20}{3}\right) = 4\left(6\left(\frac{20}{3}\right) – 110\right) = 4\left(40 – 110\right) = 4(-70) = -280$$
$$\text{Thus, } V”\left(\frac{20}{3}\right) < 0$$
$$\text{The volume has a local maximum at } x = \frac{20}{3}.$$
LAQ-7 : A window is in the shape of a rectangle surmounted by a semi-circle. If the perimeter of the window be 20 feet then find maximum area
$$\text{For the rectangle, breadth } = x, \text{ length } = 2r$$
$$\text{For the semicircle, radius } = r$$
$$\text{Given total perimeter of the window is 20 feet}$$
$$2x + 2r + \pi r = 20$$
$$2x = 20 – 2r – \pi r \quad \text{(1)}$$
$$\text{Area of the window } A = \text{Area of rectangle } + \text{Area of semicircle}$$
$$A(r) = (2r)x + \frac{\pi r^2}{2} = r(2x) + \frac{\pi r^2}{2}$$
$$A(r) = r(20 – 2r – \pi r) + \frac{\pi r^2}{2}$$
$$= 20r – 2r^2 – \pi r^2 + \frac{\pi r^2}{2} = 20r – r^2(2 + \pi – \frac{\pi}{2})$$
$$= 20r – r^2(4 + \frac{\pi}{2}) \quad \text{(2)}$$
$$A'(r) = 20 – 2r(4 + \frac{\pi}{2}) = 20 – r(4 + \pi) \quad \text{(3)}$$
$$20 – r(4 + \pi) = 0$$
$$r(4 + \pi) = 20$$
$$r = \frac{20}{4 + \pi}$$
$$A”(r) = – (4 + \pi) \quad \text{(4)}$$
$$\text{Thus, } A”(r) < 0 \text{ for all } r$$
$$A = 20\left(\frac{20}{4 + \pi}\right) – \left(\frac{20}{4 + \pi}\right)^2\left(4 + \frac{\pi}{2}\right)$$
$$= \left(400/(4 + \pi)\right) – \left(400/(4 + \pi)^2\right)\left(4 + \frac{\pi}{2}\right)$$
$$= 400/(4 + \pi) – 200/(4 + \pi)$$
$$= 200/(4 + \pi) \text{ sq.ft}$$
LAQ-8 : A wire of length l is cut into two parts which are bent respectively in the form of a square and a circle. What are the lengths of pieces of wire so that the sum of areas is least?
$$\text{Let ‘x part’ of } l \text{ be cut into a square of side } y$$
$$\text{Perimeter of the square } 4y = x$$
$$y = \frac{x}{4} \quad \text{(1a)}$$
$$\text{Remaining part } l – x \text{ is made into a circle of radius } r$$
$$\text{Circumference } 2\pi r = (l – x)$$
$$r = \frac{l – x}{2\pi} \quad \text{(1b)}$$
$$\text{Sum of areas of square and circle is}$$
$$A(x) = y^2 + \pi r^2 = \frac{x^2}{16} + \pi \left(\frac{l – x}{2\pi}\right)^2$$
$$A(x) = \frac{x^2}{16} + \frac{(l – x)^2}{4\pi} \quad \text{(2)}$$
$$A'(x) = \frac{2}{16} x + \frac{2(l – x)}{4\pi} (-1) = \frac{x}{8} – \frac{(l – x)}{2\pi} \quad \text{(3)}$$
$$\frac{x}{8} – \frac{(l – x)}{2\pi} = 0$$
$$\frac{x}{8} = \frac{(l – x)}{2\pi}$$
$$\frac{x}{4} = \frac{(l – x)}{\pi}$$
$$x\pi = 4(l – x)$$
$$x\pi = 4l – 4x$$
$$x(\pi + 4) = 4l$$
$$x = \frac{4l}{\pi + 4}$$
$$l – x = l – \frac{4l}{\pi + 4}$$
$$A”(x) = \frac{1}{8} + \frac{1}{2\pi} \quad \text{(4)}$$
$$\text{Thus, } A”(x) > 0 \text{ for all } x$$
$$\text{The total area } A(x) \text{ has a local minimum at } x = \frac{4l}{\pi + 4}.$$