VBA：関数の最小値を呼び出すにはどうすればいいですか？

Question

iとj（x、y）座標をmsgboxとして印刷するにはどうすればよいですか？

Sub Length_Min() 
    'Define variables where length and maximum lengths are integers' 
    Dim Length_Max As Double, Length As Double 

    'Comparing Variable, set maximum length to 1000m' 
    Length_Max = 1000 

    'Double for loop, within domain with increment of 0.01m.' 
    'Domain x[i] : 0<=x<=50 and Domain y[j]: 0<=y<=30' 
    For i = 0 To 50 Step 0.01 
     For j = 0 To 30 Step 0.01 

      'Subsitute x-values [i] and y- values [j] into the Total Length equation, Length.' 
      Length = (Sqr((5 - i)^2 + (5 - j)^2)) + (Sqr((10 - i)^2 + (20 - j)^2)) + (Sqr((30 - i)^2 + (10 - j)^2)) 

      'If the length found from the equation Length [L(x,y)] is less than 1000, then replace Length Max with new Length found.' 
      'Repeat for all increments of 0.01m untill smallest L(x,y) is found.' 
      If Length < Length_Max Then Length_Max = Length 

     Next j 
    Next i 

    'Print the minimum length found and its (x,y) coordinates, point P' 
    MsgBox "The minimum length is : " & Length_Max & " meters" & vbNewLine & "Minimum x and y coordinates are : " & i & " , " & j 
End Sub 

Answer 1

あなたは最小が発生したときにiとjの値の記録を保持する必要があります。

Sub Length_Min() 
    'Define variables where length and maximum lengths are integers' 
    Dim Length_Max As Double, Length As Double 
    Dim iMin As Double 
    Dim jMin As Double 

    'Comparing Variable, set maximum length to 1000m' 
    Length_Max = 1000 

    'Double for loop, within domain with increment of 0.01m.' 
    'Domain x[i] : 0<=x<=50 and Domain y[j]: 0<=y<=30' 
    For i = 0 To 50 Step 0.01 
     For j = 0 To 30 Step 0.01 

      'Subsitute x-values [i] and y- values [j] into the Total Length equation, Length.' 
      Length = (Sqr((5 - i)^2 + (5 - j)^2)) + (Sqr((10 - i)^2 + (20 - j)^2)) + (Sqr((30 - i)^2 + (10 - j)^2)) 

      'If the length found from the equation Length [L(x,y)] is less than 1000, then replace Length Max with new Length found.' 
      'Repeat for all increments of 0.01m untill smallest L(x,y) is found.' 
      If Length < Length_Max Then 
       Length_Max = Length 
       iMin = i 
       jMin = j 
      End If 
     Next j 
    Next i 

    'Print the minimum length found and its (x,y) coordinates, point P' 
    MsgBox "The minimum length is : " & Length_Max & " meters" & vbNewLine & "Minimum x and y coordinates are : " & iMin & " , " & jMin 
End Sub