Math problem

(a) Let C = K1V + K2A where K's are constants
Then:
C = K1x3 + K2x2
When x = 3:
324 = 27K1 + 9K2
When x = 4:
608 = 64K1 + 16K2
Solving, we have:
K1 = 2
K2 = 30
Therefore, C = 2x3 + 30x2
(b) Sub x = 5 into the expression:
C = 2(5)3 + 30(5)2
= 1000
So the cost for making a cube of length 5m is $1000.

(a) C = xV + yA
324 = x*(3*3*3) + y*[(3*3)*6]
324 = 27x + 54y
12 = x + 2y
x = 12 - 2y

608 = x*(4*4*4) + y*[(4*4)*6]
608 = 64x + 96y
608 = 64(12 - 2y) + 96y
608 = 768 - 128y + 96y
-160 = -32y
y = 5
if y=5, x = 12 - 2(5)
x = 2
C = 2V + 5A

(b) C = 2(5*5*5) + 5*[(5*5)*6]
= 1000

a)

324=3X3X3V+6X3X3A
12=V+2A-----------------1

608=4X4X4V+6X4X4A
608=64V+96A
19=2V+3A-----------------2

b)

24=2V+4A-------------1
19=2V+3A-------------2

1-2

A=5
V=2

C=5X5X5V+6X5X5A
C=250+750
C=1000

(a)

C = k₁V + k₂A
For a cube of length 3 m, A = 6 x 3² = 54 m², V = 3³ = 27 m³, C = 324
324 = 27 k₁+ 54 k₂ ---- (1)
For a cube of length 4 m, A = 6 x 4² = 96 m², V = 4³ = 64 m³, C = 608
608 = 64 k₁+ 96 k₂ ---- (2)
(1) x 64 - (2) x 27,
4320 = 864k₂
k₂= 5
Put k₂= 5 into (1), we get k₁= 2
Hence, C = 2V + 5A

(b)

the cost for making a cube of length 5 m
= 2 x 5³ + 5 x (6 x 5²)
= $ 1000