Find two positive numbers whose product is 100 and whose sum is a minimum.

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Tess Pollard · Accepted Answer

Step 1It is given that the difference of two numbers is 100.x-smallery-largerxy=100y=100xWe write a function that represents the minimum sum of two numbers.f(x,y)=x+ySubstitute y=100+x into f(x,y)=x+y.f(x)=x+100x=100x+100xNow, we find the first derivative of the function f(x)=x+100x.f′(x)=(x+100x)′=(x)′+(100x)′=1−100x2Step 2Find the critical points. Solve the equation f′(x)=0.f′(x)=1−100x20=1−100x2−1=−100x2x2=100x=100=10x=−100=−10 Now, find the second derivative of the function f(x).Use the second derivative test to determine whether the number we found was a critical number.f″(x)=(1−100x2)′=(1)′−(100x2)′=200x3Calculate f″(10).f″(10)=2001000=0.2&amp;gt;0Since, it is positive, it means that yes, there is a minimum.Calculate y. Substitute x=10 ∫o xy=100.10y=100y=10

cieloeventosm4 · Accepted Answer

Let the numbers be x,yxy=100,y=100xf(x,y)=x+yf(x)=x+100xMinimum of f(x) is obtained at f′(x)=0⇒f′(x)=1−100x2=0x=±10∴x=10,y=10

Find two positive numbers whose product is 100 and whose sum is a minimum.

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