Recent content by gevni

I think I batter had to ask it as a separate question (with detail) as it is a different problem. Many thanks for your help :)

Thank you I compared points closed by and get to know the minimum. Now there is one more thing, I have to provide the range of x and y within which if their values fall then there is a solution else there will be no solution. Can you please look into the attached figure it is n*n square divided...

Thank you! yes after getting partial derivatives with x and y I set the resulting equations to 0 for getting values of x and y and got the formula for x= ...and y=... Just want to confirm that now I have to replace x and y in original cost equation with these formulas (x= and y=) and it will...

I have a formula for cost calculation that contain x and y two variable. I have to find the value of (x,y) where that formula will gives minimum value as cost should not be equal to zero, it has some minimum value. I took 1st partial derivative with respect to x and then with y and found the...

yes

Thank you soooo much for helping. I was confused about whether I have to use $y' = 0$ at $x = \sqrt{\dfrac{CT}{A-B}}$ or $y' = 0$ at $x = - \sqrt{\dfrac{CT}{A-B}}$ And your explanation made it easy to understand.

Ok! so I have a formula for cost calculation. I have to find the value of x where that formula will gives minimum value as cost should not be equal to zero, it has some minimum value. The original formula is \[ cost= A\times x+ B \times(n-x)+ C\times \frac{T}{x} \] As I am only interested in...

Hi, I am trying to find the minimum root (x) of one formula. For that, I took 2nd derivative and got this equation. \[ 2 \times A \times (\frac{T}{x^3})=0 \] Here A,T,x are greater then 0. I don't know how to proceed further, how to solve it for x? Can you please guide me?

Sorry about the confusion! Let me re-write my problem: $$ {GIVEN } \\ 0 < x \le 5 \implies 0 < f(x) \le 5 \text { and } f(x) \text { is continuous,}\\ and \\ 5 \le x < 10 \implies 5 \le g(x) < 10 \text { and } g(x) \text { is continuous,}\\ $$ if f(x) or g(x) are not in the range then I am not...

I have 2 quadratic functions and I am interested in their root in the specific range. I use quadratic equation to get their roots and what I find that if their any real solution exist for both or any of the function that lie in it designated specific range, then the roots are maximum or minimum...

Might be I am asking a silly question but really want to clarify that would critical points and roots are same terms use interchangeably? I mean we can use critical point as value of x and root also as value of x then what is the difference between?

Thanks Mark! As I am interested in extrema points so I took 2nd derivative not the first 1 and simplify it into cubic equation of degree 3. And then using Cardano’s formula I found the roots and took real root as my minimum value of x.

Thank you for clarifying it.

No problem but can you please let me know that the same formula that was shown for \[ ax^3+bx^2+cx+d=0 \] we can use for \[ ax^3+bx^2-cx+d=0 \]

Thank you for reply. The part of your post is not displayed which shows the derivation steps. Can you please post them again