Answer:
[tex]60, 6[/tex]
Step-by-step explanation:
Let [tex]x[/tex] be one number and [tex]y[/tex] be the other number.
The product of the two numbers is 360.
[tex]xy=360\\\Rightarrow y=\dfrac{360}{x}[/tex]
Sum of one of the numbers and 10 times the other number be z
[tex]z=x+10y\\\Rightarrow z=x+10\times\dfrac{360}{x}\\\Rightarrow z=x+\dfrac{3600}{x}\\\Rightarrow z=\dfrac{x^2+3600}{x}[/tex]
Differentiating with respect to [tex]x[/tex] we get
[tex]z'=\dfrac{2xx-(x^2+3600)1}{x^2}\\\Rightarrow z'=\dfrac{2x^2-x^2-3600}{x^2}\\\Rightarrow z'=\dfrac{x^2-3600}{x^2}[/tex]
Equating with zero
[tex]0=\dfrac{x^2-3600}{x^2}\\\Rightarrow x=\sqrt{3600}\\\Rightarrow x=60[/tex]
Double derivative of z
[tex]z''=\dfrac{2xx^2-2x(x^2-3600)}{x^4}\\\Rightarrow z''=\dfrac{2x^3-2x^3-7200x}{x^4}\\\Rightarrow z''=\dfrac{7200}{x^3}[/tex]
at [tex]x=60[/tex]
[tex]z''=\dfrac{7200}{60^3}\\\Rightarrow z''=0.033>0[/tex]
So [tex]z[/tex] is minimum at [tex]x=60[/tex]
[tex]y=\dfrac{360}{x}\\\Rightarrow y=\dfrac{360}{60}\\\Rightarrow y=6[/tex]
So, one number is 60 and the other number is 6.
The positive numbers are 10 and 60
Let the numbers be x and y, such that x is the minimum.
So, we have:
[tex]x \times y = 360[/tex] --- the product of both numbers
[tex]y = 10x[/tex] --- the bigger number is 10 times the smaller number
Substitute 10x for y in [tex]x \times y = 360[/tex]
[tex]x \times 10x = 360[/tex]
This gives
[tex]10x^2 = 360[/tex]
Divide both sides by 10
[tex]x^2 = 36[/tex]
Take positive square roots of both sides
[tex]x = 6[/tex]
Substitute 6 for x in [tex]y = 10x[/tex]
[tex]y = 10 \times6[/tex]
[tex]y = 60[/tex]
Hence, the positive numbers are 10 and 60
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