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  1. bigroo84

    New to Mathcad

    Thanks excognito for all of your help & I will look at your notes when I get home. You have been a great help. jghrist that is an interesting use of the first derivative root for me.
  2. bigroo84

    New to Mathcad

    Thank you, I'm looking into the points you raised. I've attached a graph of the function V(x). The maximum value on it shows 14.993. Can a 2nd derivative test prove this do you think?http://files.engineering.com/getfile.aspx?folder=73dc5224-aa4f-48db-9d11-91124ef669b6&file=Engtips_1.mcd
  3. bigroo84

    New to Mathcad

    Right I was okay up until the last bit & these are probably daft questions but you have helped me immensely so far: What is the answer to the max volume as there are 3 values? -10 10.657 (zero y value)(root) -0.657 (zero y value)(root) When you write "V(roots[0)=" what is the [ for? Also...
  4. bigroo84

    New to Mathcad

    So the total volume was V(x)=(10+x)/6.sqrt(32-(5-x)^2) with x between 0 & 10.657, how do you get the maximum volume? Do I plug in values for x within the range given?
  5. bigroo84

    New to Mathcad

    Wow again! I'm learning so much just in 1 question! Why isn't a solve block suitable here? So do I understand you that roots returns a value of x when the f(x)=0 Im is also a new operator to me.
  6. bigroo84

    New to Mathcad

    Okay so if I wanted to find a value of x where V'(x)=0 I presume I use a solve block. If so would I now write: dVs2(x):=0 Given then put the dVs2(x) equal to something?
  7. bigroo84

    New to Mathcad

    Wow that sure does look a lot neater & 'simplified'. What difference does the word 'simplify' do compared to Ctrl.
  8. bigroo84

    New to Mathcad

    Hi, Can anyone take a look at my file to see if I have understood the what is being asked of me properly as I have only just started Mathcad & need all the help I can get! Thank you

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