## Friday, June 24, 2011

### Y10 Add Maths Max/min Quiz

A cuboid has a total surface area of 120 cm2. Its base measures x cm by 2x cm and its height is h cm.

i) Obtain an expression for h in terms of x.                                                                                                     

Make a sketch:

The area of the base is x × 2x = 2x2.  The top face is the same.
The area of the right side is 2x × h = 2xh. The left face is the same.
The area of the front face is x × h = xh.  The back is the same.
Surface area = 120 = 4x2 + 6xh
120 – 4x2 = 6xh
h = (120 – 4x2)/6x
h = 20x-1 – 2x/3

Given that the volume of the cuboid is Vcm3,
ii) show that V = 40x 4 x3/3 .                                                                                                               

V = x × 2x × h
V = 2 x2 (20x-1 – 2x/3)
V = 40x - 4 x3/3

Given that x can vary,
iii) show that V has a stationary value when h = 4x/3 .                                                                            
dV/dx = 40 - 4x2
0 = 40 - 4x2
x = +√10  (reject the negative root)
Substitute this value of x into h = 20x-1 – 2x/3
h = 20/√10 - 2√10/3 = 2 √10√10/√10 - 2 √10/3    (Since 20 = 2 × 10 and 10 itself = 102)
h = 2√10 - 2√10/3
h = 6√10/3 - 2√10/3
h = 4√10/3
h = 4x/3      (Recall that x = √10).

iv) Find the stationary value of V.                                                                                                       
V = 40x - 4 x3/3
= 40√10 - 4 √10 3/3               ( √10) 3  = √10√10√10 = 10√10 )
= 40(√10 - √10)/3)
= 80√10/3

iv) Determine the nature of the stationary value. Show working.                                                          
d2V/dx2 = -8x = (negative)(positive) = (negative).    Note that x itself must be > 0, i.e., positive
The function is concave down. The turning point is a maximum.