# RD Sharma Solutions For Class 12 Maths Exercise 18.5 Chapter 18 Maxima and Minima

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Exercise 18.4 Solutions

### Exercise 18.5 Page No: 18.72

1. Determine two positive numbers whose sum is 15 and the sum of whose squares is minimum.

Solution:

Which implies S is minimum when a = 15/2 and b = 15/2.

2. Divide 64 into two parts such that the sum of the cubes of two parts is minimum.

Solution:

Let the two positive numbers be a and b.

Given a + b = 64 â€¦ (1)

We have, a3Â + b3Â is minima

Assume, S = a3Â + b3

(From equation 1)

S = a3Â + (64 â€“ a)3

of S

Hence, the two number will be 32 and 32.

3. How should we choose two numbers, each greater than or equal to â€“2, whose sum is Â½ so that the sum of the first and the cube of the second is minimum?

Solution:

dS/da = 0

1 + 3(Â½ – a)2 (-1) = 0

1 – 3(Â½ – a)2 = 0

4. Divide 15 into two parts such that the square of one multiplied with the cube of the other is minimum.

Solution:

Let the given two numbers be x and y. Then,

x + y = 15 â€¦.. (1)

y = (15 – x)

Now we have, z = x2 y3

z = x2 (15 â€“ x)3 (from equation 1)

5. Of all the closed cylindrical cans (right circular), which enclose a given volume of 100 cm3, which has the minimum surface area?

Solution:

Let r and h be the radius and height of the cylinder, respectively. Then,

Volume (V) of the cylinder = Ï€r2 h

âŸ¹ 100 = Ï€r2 h

âŸ¹ h = 100/ Ï€r2

Surface area (S) of the cylinder = 2 Ï€r2 + 2 Ï€r h = 2 Ï€r2 + 2 Ï€r Ã— 100/ Ï€r2

6. A beam is supported at the two ends and is uniformly loaded. The bending moment M at a distance x from one end is given by

Find the point at which M is maximum in each case.

Solution:

7. A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the lengths of the two pieces so that the combined area of the circle and the square is minimum?

Solution:

Suppose the given wire, which is to be made into a square and a circle, is cut into two pieces of length x and y m respectively. Then,

x + y = 28 â‡’ y = (28 â€“ x)

We know that perimeter of square, 4 (side) = x

Side = x/4

Area of square = (x/4)2 = x2/16

Circumference of circle, 2 Ï€ r = y

r = y/ 2 Ï€

8. A wire of length 20 m is to be cut into two pieces. One of the pieces will be bent into shape of a square and the other into shape of an equilateral triangle. Where the wire should be cut so that the sum of the areas of the square and triangle is minimum?

Solution:

Suppose the wire, which is to be made into a square and a triangle, is cut into two pieces of length x and y respectively. Then,

x + y = 20 â‡’ y = (20 – x) â€¦â€¦ (1)

We know that perimeter of square, 4 (side) = x

Side = x/4

Area of square = (x/4)2 = x2/16

Again we know that perimeter of triangle, 3 (side) = y.

Side = y/3

Hence, the wire of length 20 m should be cut into two pieces of lengths

9. Given the sum of the perimeters of a square and a circle, show that the sum of their areas is least when one side of the square is equal to diameter of the circle.

Solution:

Let us say the sum of perimeter of square and circumference of circle be L

Given sum of the perimeters of a square and a circle.

Assuming, side of square = a and radius of circle = r

Then, L = 4a + 2Ï€r â‡’ a = (L – 2Ï€r)/4â€¦ (1)

Let the sum of area of square and circle be S

So, S = a2Â + Ï€r2

10. Find the largest possible area of a right angled triangle whose hypotenuse is 5 cm long.

Solution:

11. Two sides of a triangle have lengths â€˜aâ€™ and â€˜bâ€™ and the angle between them is Î¸. What value of Î¸ will maximize the area of the triangle? Find the maximum area of the triangle also.

Solution:

It is given that two sides of a triangle have lengths a and b and the angle between them is Î¸.

Let the area of triangle be A

12. A square piece of tin of side 18 cm is to be made into a box without top by cutting a square from each corner and folding up the flaps to form a box. What should be the side of the square to be cut off so that the volume of the box is maximum? Also, find this maximum volume

Solution:

Given side length of big square is 18 cm

Let the side length of each small square be a.

If by cutting a square from each corner and folding up the flaps we will get a cuboidal box with

Length, L = 18 â€“ 2a

Breadth, B = 18 â€“ 2a and

Height, H = a

Assuming, volume of box, V = LBH = a (18 – 2a)2

Condition for maxima and minima is

13. A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off squares from each corners and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum possible?

Solution:

Given length of rectangle sheet = 45 cm

Breath of rectangle sheet = 24 cm

Let the side length of each small square be a.

If by cutting a square from each corner and folding up the flaps we will get a cuboidal box with

Length, L = 45 â€“ 2a

Breadth, B = 24 â€“ 2a and

Height, H = a

Assuming, volume of box, V = LBH = (45 – 2a)(24 – 2a)(a)

Condition for maxima and minima is

(45 – 2a) (24 – 2a) + (- 2) (24 – 2a) (a) + (45 – 2a) (- 2)(a) = 0

4a2Â â€“ 138a + 1080 + 4a2Â â€“ 48a + 4a2Â â€“ 90a = 0

12a2Â â€“ 276a + 1080= 0

a2Â â€“ 23a + 90= 0

a = 5, 18

14. 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 square metre for the base and Rs 45 per square metre for sides, what is the cost of least expensive tank?

Solution:

Let the length, breath and height of tank be l, b and h respectively.

Also, assume volume of tank as V

h = 2 m (given)

V = 8 m3

l b h = 8

2lb = 8 (given)

lb = 4

b =Â 4/lÂ â€¦ (1)

Cost for building base = Rs 70/m2

Cost for building sides = Rs 45/m2

Cost for building the tank, C = Cost for base + cost for sides

C = lb Ã— 70 + 2(l + b) h Ã— 45

C = l(4/l) Ã— 70 + 2(l + 4/l) (2) Ã— 45 [Using (1)]

= 280 + 180 (l + 4/l)

C = Rs 1000

15. A window in the form of a rectangle is surmounted by a semi-circular opening. The total perimeter of the window is 10 m. Find the dimensions of the rectangular part of the window to admit maximum light through the whole opening.

Solution:

Let the radius of semicircle, length and breadth of rectangle be r, x and y respectively

AE = r

AB = x = 2r (semicircle is mounted over rectangle) â€¦1

Given Perimeter of window = 10 m

x + 2y + Ï€r = 10

2r + 2y + Ï€r = 10

2y = 10 â€“ (Ï€ + 2).r

16. A large window has the shape of a rectangle surmounted by an equilateral triangle. If the perimeter of the window is 12 metres find the dimensions of the rectangle that will produce the largest area of the window.

Solution:

Let the dimensions of the rectangle be x and y.

Therefore, the perimeter of window = x + y + x + x + y = 12

3x + 2y = 12

y = (12 â€“ 3x)/2 â€¦. (1)

Now,

17. Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R isÂ 2R/âˆš3.

Solution:

18. A rectangle is inscribed in a semi-circle of radius r with one of its sides on diameter of semi-circle. Find the dimensions of the rectangle so that its area is maximum. Find also the area.

Solution:

Let the length and breadth of rectangle ABCD be 2x and y respectively

Radius of semicircle = r (given)

In triangle OBA, where is the centre of the circle and mid-point of the side AC

r2Â = x2Â + y2Â (Pythagoras theorem)

y2Â = r2Â – x2

Â â€¦ (1)

Let us say, area of rectangle = A = xy

[Since, l = 2x]

19. Prove that a conical tent of given capacity will require the least amount of canvas when the height is âˆš2 times the radius of the base.

Solution:

20. Show that the cone of the greatest volume which can be inscribed in a given sphere has an altitude equal to 2/3 of the diameter of the sphere.

Solution:

Let the radius and height of cone be r and h respectively

R2Â = r2Â + (h – R)2

R2Â = r2Â + h2Â + R2Â – 2hR

r2Â = 2hR – h2Â â€¦ (1)

Assuming volume of cone be V

21. Prove that the semi – vertical angle of the right circular cone of given volume and least curved surface isÂ cot-1âˆš2

Solution:

22. An isosceles triangle of vertical angle 2Î¸ is inscribed in a circle of radius a. Show that the area of the triangle is maximum when Î¸ = Ï€/6.

Solution:

Î” ABC is an isosceles triangle such that AB = AC.

The vertical angleÂ BAC = 2Î¸

Triangle is inscribed in the circle with center O and radius a.

Draw AM perpendicular to BC.

Since, Î” ABC is an isosceles triangle, the circumcenter of the circle will lie on the perpendicular from A to BC.

Let O be the circumcenter.

BOC = 2 Ã— 2Î¸ = 4Î¸ (Using central angle theorem)

COM = 2Î¸ (Since, Î” OMB and Î” OMC are congruent triangles)

OA = OB = OC = a (radius of the circle)

In Î” OMC,

CM = asin2Î¸

OM = acos2Î¸

BC = 2CM (Perpendicular from the center bisects the chord)

BC = 2asin2Î¸

Height of Î” ABC = AM = AO + OM

AM = a + acos2Î¸

23. Prove that the least perimeter of an isosceles triangle in which a circle of radius r can be inscribed is 6âˆš3r.

Solution:

QR at X and PR at Z.

OZ, OX, OY are perpendicular to the sides PR, QR, PQ.

Here PQR is an isosceles triangle with sides PQ = PR and also from the figure,

â‡’Â PY = PZ = x

â‡’Â YQ = QX = XR = RZ = y

From the figure we can see that,

â‡’Â Area (Î”PQR) = Area (Î”POR) + Area (Î”POQ) + Area (Î”QOR)

â‡’Â PER = 2(âˆš3r) + 4(âˆš3r)

â‡’Â PER = 6âˆš3r

âˆ´Â Thus proved

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