10 (b): Budget line will be C2=M2+.75(M1-C1)
Solve it to get: (3/4)C1+C2=900----(i)
U(C1,C2)= C1.C2
We'll use the optimizing condition: MRS=-P1/P2
where MRS=-C2/C1
So this gives us: C2/C1=3/4----(ii)
From (i) and (ii), we get:
C1=600 and C2=450
And the rats will eat .25(M1-C1)= 1/4*(1000-600)=100 Bushels

10.(c). Now trade with the rotw is possible, so instead of saving this year they can trade the extra in the market and then save the money receipts at interest rate of 10%. So the Budget line (BL) would be: C1+C2/(1+r)=M1+M2/(1+r)
Solving it after plugging the given values, get:
BL: C1+C2/1.1=1136

Solving just like in previous part using MRS=-P1/P2 we get:
C1=568 and C2=624
Hope that helps :)

I'm not so sure about this part! One other thing that I don't understand is why are there kinks in the Blue and Black Budget lines in the graph that is given in the solutions book!!??-Below is the picture attached! Can someone please explain the reason for the kinks??

Sonia look at the 2nd equation on page 184 of Varian (if you have 7th edition of the book!). Also since here its given rats will eat 25% of what is stored in first year, so what is stored in first year can be thought of as the extra over what is consumed in first period I,e. (M1-C1) where M1 is the endowment in period 1. Now 25% of it is eaten by rats so what is left from the savings is .75(M1-C1) which can be had in the 2nd period so its added to M2 (endowment in 2nd period!)
So, what they can consume in 2nd period is given by C2=M2+.75(M1-C1). Hope that makes sense!!?