Respuesta :
Let xy = 25, x and y > 0.
[tex]y = \frac{25}{x}[/tex]
[tex]\text{Denote S as: } x + y[/tex]
[tex]S(x) = x + \frac{25}{x}[/tex]
[tex]\text{Minimum: }S'(x) = 0, S''(x) > 0[/tex]
[tex]S'(x) = 1 - \frac{25}{x^{2}}[/tex]
[tex]1 - \frac{25}{x^{2}} = 0[/tex]
[tex]x^{2} = 25[/tex]
[tex]x = 5 (x > 0)[/tex]
[tex]S''(x) = \frac{50}{x^{3}} > 0, (x > 0)[/tex]
Thus, at x = 5, there lies a minimum summation, and this forms a square with side length of 5.
[tex]y = \frac{25}{x}[/tex]
[tex]\text{Denote S as: } x + y[/tex]
[tex]S(x) = x + \frac{25}{x}[/tex]
[tex]\text{Minimum: }S'(x) = 0, S''(x) > 0[/tex]
[tex]S'(x) = 1 - \frac{25}{x^{2}}[/tex]
[tex]1 - \frac{25}{x^{2}} = 0[/tex]
[tex]x^{2} = 25[/tex]
[tex]x = 5 (x > 0)[/tex]
[tex]S''(x) = \frac{50}{x^{3}} > 0, (x > 0)[/tex]
Thus, at x = 5, there lies a minimum summation, and this forms a square with side length of 5.
The two positive values whose product is 25 and their sum is a minimum are; +5 and +5.
Let the two numbers be x and y.
We are told that their product is 25
Thus;
xy = 25
y = 25/x
We are told their sum is a minimum.
Thus;
S = x + y
Putting 25/x for y gives;
S = x + (25/x)
Since the sum is minimum, let us find the derivative and equate to zero;
S'(x) = 1 - 25/x²
At S'(x) = 0, we have;
1 - 25/x² = 0
25/x² = 1
x² = 25
Thus;
x = √25
x = 5
Putting 5 for x into y = 25/x;
y = 25/5
y = 5
Thus, the two positive values are 5.
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