6.
a.
CMx=220-70-2x30-1.5x20=$60
CMy=350-110-4x30-1x20=$100

b.
Objective: Maximize 60X+100Y
Constraints:
2X+4Y≤12,000 1.5X+Y≤6,000 X≤3,500

Using Solver, we can find the optimal mix of X and Y is: X=3,000; Y=1,500.
The corresponding optimal CM=$330,000

The solver output is shown below:

c. The graph above consists all the constraints functions of this case. When x=3,000 the objective function achieves optimal solution. The optimal solution can only be achieved when the objective function’s slope is between 0.5 (the slope of constraint function 2x+4y=12,000) and 1.5 (the slope of constraint function 1.5x+y=6,000). 60CMy=1.5→CMyminimum=40 60CMy=0.5→CMymaximum= 120
Thus, contribution margin of Y can increase by $20 (=120-100) or decrease by $60 (=100-40) without changing the optimal product mix.

10.
a.
Investment: I
Revenue: R for both years
Depreciation: {d, 1-d}

RI1=R-dI-rI=R-I(d+r)
RI2=R-(1-d)I-rI(1-d)=R-I(1-d)(1+r)

PV(RI)=R-I(d+r)1+r+R-I(1-d)(1+r)(1+r)2=R1+r-Id+r1+r+R1+r2-I1-d1+r1+r2=R1+r+R1+r2-Id+Ir1+r+I-Id1+r1+r2=R1+r+R1+r2-Ir2+2Ir+I1+r2=R1+r+R1+r2-Ir+121+r2=R1+r+R1+r2-I=NPV

b.
In order to solve for the annuity depreciation rate d* we need to ensure that RI1=RI2.
Or,
R-dI-kI=R-(1-d)I-kI(1-d)
-dI-kI=-I+dI-kI+kId
d(-2I-kI)=-I d*=I2I+kI=12+r . Since there is not R or I in the notation of d*, d* is not relevant to R or I.
Thus, the owner should use depreciation schedule {12+r , 12+r}. The depreciation schedule is not relevant to R or I.

c. d*1-d*=12+r1-12+r=12+r2+r-12+r=12+rr+12+r=1r+1 . Since r is a positive rate, 1r+1<1. Thus, d*<1-d*. This means in Year 2 depreciation is higher.

12.
a. I-0.11+r-I=0→ I-0.11+r=I→r=I-0.1I-1
In order to maximize IRR, we need to take derivative in terms of I for the above function.