Please help me to solve those two problems by using mathematica.(please show me
ID: 3167505 • Letter: P
Question
Please help me to solve those two problems by using mathematica.(please show me the steps.)
problem 1
Let x be the following consumption bundleLet x be the following consumption bundle
x = {0.2653087455598462`, 0.9191829435531803`, 0.637298689661289`,
0.9635628224868256`, 0.5003059294177169`, 0.5192334096822956`,
0.7035429296939456`, 0.6280771321481432`, 0.8245868402598907`,
0.8639345189381371`, 0.7314564373632568`, 0.5157765921979305`,
0.8293122011630816`, 0.0215316690843516`, 0.8481298029638905`,
0.6287682785332761`, 0.3951971283324276`, 0.7022145202119485`,
0.9479829321230706`, 0.13662270412038577`};
and p be the following price vector
p = {0.8031401722985496`, 0.3219824373217741`, 0.7249113715261456`,
0.6604168661242432`, 0.5378314267387034`, 0.40279949376859375`,
0.0876126818648566`, 0.6968540436374175`, 0.0375254973209865`,
0.8835660840862981`, 0.38406975217091105`, 0.06877691148927442`,
0.21293865706109585`, 0.019631565148161088`, 0.6526133148076542`,
0.553000319291344`, 0.3836264558980143`, 0.9980998960638094`,
0.8044835118437638`, 0.9242320407580679`};
If a consumer's income is 5, can she afford to buy this bundle?
Problem 2
Let x,y, and z be following three vectors in [DoubleStruckCapitalR]^10:
x = {3, -4, 1, 0, 2, 3, 0, -4, -1, -1};
y = {-4, 5, 2, -4, -1, -1, -4, 1, 5, -1};
z = {4, 3, -1, 4, -1, 4, -1, -1, -3, 3};
(i) Compute the distance between x and y; the distance between y and z; and the distance between x and z. Verify the triangle inequality d(x,z)<= d(x,y)+d(y,z).
ii) Are the points y and z in the open ball B(x,1/2)?
Thanks alot
Explanation / Answer
Answer 1:
Take the dot product of two vectors
res = x.p
if res <= 5 consumer can afford the bundle otherwise not
Ans 2:
Ans i:
distance between x & y = EuclideanDistance(x,y)
distance between y & z = EuclideanDistance(y,z)
distance between z & x = EuclideanDistance(z,x)
Ans ii :
EuclideanDistance(z,x) < 1/2 then z is in B(x,1/2)
EuclideanDistance(x,y) < 1/2 then y is in B(x,1/2)