2 mathematical questions?

Question

a) A rectangular boz is to be made fomr a piece of cardboard 6cm wide and 14 cm long. It will be formed by cutting out squares from each of the four corners, and then turning up the sides. If the volume of hte box is to be 40 cubic cm, what should be the dimensions of the squares that are cut out?

b) For the same box, what if we don't have a particular vloume in mind but we wnat the volume of the box to be as large as possible, then what size squares do we cut out?

Answer 1

a) The width of the boz will be 6 - 2x, the length of the boz will be 14 - 2x, and the height of the boz will be x.

x(6 - 2x)(14 - 2x) = 40
x(84 - 40x + 4x^2) = 40
4x^3 - 40x^2 + 84x - 40 = 0
x^3 - 10x^2 + 21x - 10 = 0

Answer 2

V = x(6 - 2x)(14 - 2x) = 40
84x - 12x^2 -28x^2 + 4x^3 = 40
x^3 - 10x^2 + 21x - 10 = 0
(x - 2)(x^2 - 8x - 5) = 0
x = (8 ± √(64 + 20))/2
x = 4 ± √(16 + 5)
x = 4 ± √21
x = 2 (neither of the irrational roots works)

Checking:
2(6 - 4)(14 - 4) =?40
2*2*10 = 40

for max volume,
V = 4x^3 - 40x^2 + 84x
12x^2 - 80x + 84 = 0
3x^2 - 20x + 21 = 0
x = (20 ± √400 - 252))/6
x = (10 ± √100 - 63))/6
x = (10 - √37)/3
x = 1.3057 cm, or 13 mm
Then
V = 1.3057(6 - 2*1.3057)(14 - 2*1.3057)
V = 50.3888 cm^3

Answer 3

confident!!! a^2 + a^2 = 2a^2 it rather is like asserting 3^2 + 3^2 = 2 x 3^2 = 18 yet in a diverse way is to factorise a^2 + a^2 = a(a + a) = a(2a) = 2a^2 In Numbers 5^2 + 5^2 = 5(5 + 5 ) = 5 x 10 = 50 5^2 + 5^2 = 25 + 25 = 50

Answer 4

2x2.

b. wont matter.