The linear relationship shows constant slope, i.e., Y = a + bX
dY/dX = b
(where "d" means "changes in")

i) Quantity Profit
14 $7,000
0 $0
19 $9,000
11 $5,500

When Q changes from 0 to 14, dQ = 14, and dProfit = 7000
Then dProfit/dQ = 7000/14 = 500
However, from 14 to 19, dQ = 19-14 = 5, but dProfit = 9000-7000 = 2000
Then dProfit/dQ = 2000/5 = 400
So the slope is not constant, the relationship is NOT linear.

ii) Quantity Metal Used
7 17.5 oz
0 0 oz
9 23.5 oz
3 7.6 oz
When Q changes from 0 to 3, dQ = 3, and dY = 7.6
Then dY/dQ = 7.6/3 = 2.533
However, from 3 to 7, dQ = 7-3 = 4, but dY = 17.5 - 7.6 = 9.9
Then dY/dQ = 9.9/4 = 2.475
So the slope is not constant, the relationship is NOT linear.

iii) Quantity Fxed Expeneses
3 oz $4,200
0 oz $4,200
5 oz $4,200
1 oz $4,200
No matter what is the change in Q, the expense is always 4200, which means the slope is always zero.
However, there is no relationship between Q and expenses.

iv) Quantity Labor Used
0 0 hrs
7 52.5 hrs
10 75 hrs
2 15 hrs

When Q changes from 0 to 2, dQ = 2, and dL = 15
Then dL/dQ = 15/2 = 7.5
When Q changes from 2 to 7, dQ = 7-2 = 5, and dL = 52.5 - 15 = 37.5
Then dL/dQ = 37.5/5 = 7.5
When Q changes from 7 to 10, dQ = 10-7= 3, and dL = 75-52.5= 22.5
Then dL/dQ = 22.5/3 = 7.5
So the slope is constant, the relationship is linear.

v) Shoes Belts
0 19
5 4
6 1
1 16
When S changes from 0 to 1, dB =1, and dB= 16-19= -3, then dB/dS = -3
When S changes from 1 to 5, dS = 4, and dB = 4 - 16 = -12, then dB/dS = -12/4 = -3
When S changes from 5 to 6, dS = 1, and dB = 1-4 = -3, then dB/dS = -3
So the slope is constant, the relationship is linear.

vi) Quantity Metal Used
10 33.0 oz
15 49.5 oz
7 23.1 oz
9 29.7 oz
When Q changes from 7 to 9, dQ = 2, and dM= 29.7-23.1= 6.6, then dM/dQ = 3.3
When S changes from 9 to 10, dQ = 1, and dM= 33.0-29.7= 3.3, then dM/dQ = 3.3
When S changes from 10 to 15, dQ = 5, and dM = 49.5-33 = 16.5, then dM/dQ = 3.3
So the slope is constant, the relationship is linear.

vii) A=B^2 + 2
By derivative calculation,
dA/dB = 2B
it varies with the value of B, thus, the slope is not constant, the relationship is NOT linear.

viii) y=x
By derivative calculation,
dy/dx = 1
the slope is constantly 1, the relationship is linear.