Find the max value of xy (done in a weird way)

  • Thread starter Thread starter Andrax
  • Start date Start date
  • Tags Tags
    Max Value Weird
Andrax
Messages
117
Reaction score
0
find the max value of xy (done in a "weird" way)

Homework Statement


so our teacher assigned this homework in the last 10 minutes
we haven't read the function with 2 variables and I've never seen/used one
(things were just getting in my head since it's only 10 mins i don't know if I'm making a mistake)
suppose that x and y satisfie 4x+y=2
find the maximum value of xy
so this is done with deriviatives but i was thinking in using another way and it turned to be right , what I'm asking here is explaining why my method was right, anyway the answer is (1/4 , 1) this is how I've done it
f(x)=4x+y-2 must be eqUAL TO 0
f(xy)=4xy+y-2 this must be equal to 0( assumed that y is not changing i don't know )
anyway this leads to x=2-y/4y
replacing in 4x+y=2 we get y= 1 then x=1/4
now our teacher used derviatives to solve this
y=2-4x
xy=2x-4x^2 etc and he gets the exact same answer as me , i don't think this is a coincidence so please explain to me why this is right , when i asked him he mentioned two variable functions..
thanks everyone :)
 
Last edited by a moderator:
Physics news on Phys.org
Andrax said:

Homework Statement


so our teacher assigned this homework in the last 10 minutes
we haven't read the function with 2 variables and I've never seen/used one
(things were just getting in my head since it's only 10 mins i don't know if I'm making a mistake)
suppose that x and y satisfie 4x+y=2
find the maximum value of xy
so this is done with deriviatives but i was thinking in using another way and it turned to be right , what I'm asking here is explaining why my method was right, anyway the answer is (1/4 , 1) this is how I've done it
f(x)=4x+y-2 must be eqUAL TO 0
f(xy)=4xy+y-2 this must be equal to 0( assumed that y is not changing i don't know )
Where did 4xy + y - 2 come from? In particular, the 4xy part.

Edit: Now I see what you did, which was to substitute xy for x in the function definition. Even so, I'm not sure that what you did after that makes sense.
Andrax said:
anyway this leads to x=2-y/4y
What you wrote is x = 2 - (y/4 * y). Is that what you meant? Or did you mean x = (2 - y)/(4y)?
Andrax said:
replacing in 4x+y=2 we get y= 1 then x=1/4
now our teacher used derviatives to solve this
y=2-4x
xy=2x-4x^2 etc and he gets the exact same answer as me , i don't think this is a coincidence so please explain to me why this is right , when i asked him he mentioned two variable functions..
thanks everyone :)
 
Last edited:
Yes it doesn't look logical anyway thanks
 
Andrax said:

Homework Statement


so our teacher assigned this homework in the last 10 minutes
we haven't read the function with 2 variables and I've never seen/used one
(things were just getting in my head since it's only 10 mins i don't know if I'm making a mistake)
suppose that x and y satisfie 4x+y=2
find the maximum value of xy
so this is done with deriviatives but i was thinking in using another way and it turned to be right , what I'm asking here is explaining why my method was right, anyway the answer is (1/4 , 1) this is how I've done it
f(x)=4x+y-2 must be eqUAL TO 0
f(xy)=4xy+y-2 this must be equal to 0( assumed that y is not changing i don't know )
This is incorrect. Because 4x+ y- 2= 0 for all x and y, if x changes, y must also change. In particular, you cannot write "f(x)= 4x+ y- 2" because f depends upon values of both x and y.

anyway this leads to x=2-y/4y
replacing in 4x+y=2 we get y= 1 then x=1/4
now our teacher used derviatives to solve this
y=2-4x
xy=2x-4x^2 etc and he gets the exact same answer as me , i don't think this is a coincidence so please explain to me why this is right , when i asked him he mentioned two variable functions..
thanks everyone :)
Looks to me like shear coincidence. Your "method" is NOT valid.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top