₦airaland Forum

Laplacian:
w= (z+2j)/(z+j), w(z+j)=z+2j or

z=(2j+wj)/(w-1)=[(u+2)j-v]/[(u-1)+vj] or
taking conjugates;
z=[(u+2)j-v]*[(u-1)-vj]/[(u-1)^2+v^2]

={[(u-1)(u+2)+v^2]j+[-uv+v(u+2)]}/[(u-1)^2+v^2]
so
x=[-uv+v(u+2)]/[(u-1)^2+v^2]
and
y=[(u-1)(u+2)+v^2]/[(u-1)^2+v^2]

if u make this substitution u 'll get the centre

agentofchange1:
greet thee all

try this plz.


Q1)Use Lagrange multiplier to find the values of x,y,z that minimises the objective function f(x,y,z) =11xy+14yz+15xz, subject to the constrain xyz=147840

Q2) A rectangular box open at the top is to have a volume of 32cm-cube , what must be the dimension of the box so that the total surface is maximum .

Laplacian:
recall, sec²y=1+tan²y.
From the LHS,
sec²x-1=1+2(sec²y-1) or
sec²x=2sec²y or
cos²y=2cos²x or
cos²y=(2cos²x-1)+1, does the expression in the bracket look familiar? *****


u can also use the condition given above to dissolve the question @ sir laplacian.....
- i used it ....


That's cos2x.
So, cos²y=cos2x+1
or
cos²y-1=cos2x
can u finish it from here? Hop it helps!

actYourDreams:
Your problem was not well-defined. Math is a concise language so when posing a math problem you need to be precise.

By saying "prove that x = 2", I will assume that your solution has to be a natural number. Hence, my following approach is based on the assumption that x must be a natural number.

Observe that 4^x = 2^(2x) and 5^x = (3+2)^x. Therefore your problem can be written as
2^(2x)+ 3^x=(3+2)^x.

Then apply binomial expansion to (3+2)^x, i.e.
(3+2)^x = 3^x + 2x3^(x-1) + (x(x-1))/2 * 3^(x-2)2^2 + ... + x*3*2^(x-1) + 2^x.

Therefore the question becomes

2^(2x) = 2x3^(x-1) + (x(x-1))/2 * 3^(x-2)2^2 + ... + x*3*2^(x-1) + 2^x, since the 3^x on both sides cancel out.

Then it easily follows that for the previous equality to hold x must be 2 or use mathematical induction to show that.

If NL had Latex capability I might have clarified further. However, Ihope my approach was clear enough.