Question

A multipurpose hall with square base wooden floor of side x and height h is to be constructed in a school with sound proof material used in four walls and in roof of area the area of material used is k square units.

Based on the above information answer the following:
1. The relation between x, h and k is:
A) k^2= 2x+4xh B) k^2= x^2+4xh C) x^2=k^2+4xh D)x^2=2k^2+2xh

2.Volume of air present in the hall:
A) x(k2-x2)/4 B)x(k2+x2)/4 C)k(k2-x2)/4 D) x(k2-x2)/4

3)The area of four walls is:

(a) k2 + x2 (b) k2 – x2 (c) x2 – k2 (d) none of these

4)The CEO of the school is interested in maximizing the volume of the hall. For this to happen the value of x should be :

(a) 3/root k (b) root k/3 (c) k/ root 3 (d) root 3/ root k

5)The maximum volume of the hall is (a) k^2/6 root 3 (b) k^3/2 root 3 (c) k^3/3 root 3
(d) k^3/ 6 root 3

Answer 1

Answer:

Since the top of the building is in the form of half of the cylinder of radius 3.5m, and length 10m, split along the diameter

V= volume of the godown

V= Volume of the cuboid +

2

1

(Volume of the cylinder of radius 3.5m and length 10m)

=(l×b×h)+

2

1

(πr

2

h)

={10×7×3+

2

1

(

7

22

×3.5×3.5×10)}m

3

=(210+192.5)m

3

=402.5m

3

total surface area excluding the base floor

=Area of four walls +

2

1

Curved surface area of the cylinder + 2 area of the semi-circles

=[2(10+7)×3×

2

1

(

7

22

×3.5×3.5×10)+2(

7

22

×3.5×3.5)]m

2

=250.5m

2

Answer 2

Answer:

1.)  option (B) k² = 4xh + x²

2.) option (D) x(k² - x²)/4

3.) option (D) none of these

4.) option (C) x = $\frac{k}{\sqrt{3} }$

5.) option (D)  $\frac{k^3}{6\sqrt{3} }$  

Step-by-step explanation:

Given length = breadth = x

height = h

Area of material used = k² sq. units

(1.)  Area of material used = Area of 4 walls + Area of roof

=> k² = 4 * xh + x*x

=> k² = 4xh + x²

Therefore, option (B) k² = 4xh + x² is correct

(2.) Volume of air = Volume of the hall = Length * Breadth * Height

=> Volume = x * x * h

                  = x²h

As the options do not contain 'h', we will substitute its value from part (1)

=> Volume = x²(k² - x²)/4x

                  = x(k² - x²)/4

Therefore, option (D) x(k² - x²)/4 is correct

(3.) Area of four walls = 4 * Area of 1 wall

                                    = 4 * length * breadth

                                    = 4 * x * x

                                    = 4x²

Therefore, option (D) none of these is correct

(4.) We need to maximize volume

Volume (V) = x(k² - x²)/4 = $\frac{1}{4}$ * (k²x - x³)

$\frac{dV}{dx}$ = $\frac{1}{4}$ * [k² - 3x²]

$\frac{dV}{dx}$ = 0

=>  $\frac{1}{4}$ * [k² - 3x²] = 0

=> k² - 3x² = 0

=> k² = 3x²

=> k = x √3

=> x = $\frac{k}{\sqrt{3} }$

$\frac{d^2V}{dx^2}$ = $\frac{1}{4}$ * [0 - 6x]

     = $\frac{-6x}{4}$ < 0 Therefore maxima at x = $\frac{k}{\sqrt{3} }$

Therefore, option (C) x = $\frac{k}{\sqrt{3} }$ is correct

(5.) Maximum volume =  $\frac{-6x}{4}$ (proved above)

Now substituting x = $\frac{k}{\sqrt{3} }$ in the volume

=> Maximum volume =  $\frac{1}{4}$ * [k²$\frac{k}{\sqrt{3} }$  - ($\frac{k}{\sqrt{3} }$ )³]

                                   =  $\frac{1}{4}$ * [$\frac{k^3}{\sqrt{3} }$  - $\frac{k^3}{3\sqrt{3} }$ ]

                                   =  $\frac{1}{4}$ * [$\frac{k^3}{\sqrt{3} }$  (1 -$\frac{1}{3}$)]

                                   =  $\frac{1}{4}$ * $\frac{k^3}{\sqrt{3} }$  * $\frac{2}{3}$

                                   =  $\frac{1}{6}$ * $\frac{k^3}{\sqrt{3} }$  

                                   =  $\frac{k^3}{6\sqrt{3} }$  

Therefore, option (D)  $\frac{k^3}{6\sqrt{3} }$  is correct