Differential equationsFind two positive numbers whose product is 100 and whose sum is a minimum.
4 weeks ago
Find two positive numbers whose product is 100 and whose sum is a minimum.
2 Answers
Best Answer
davisbearden Staff answered 4 weeks ago
Step 1
It is given that the difference of two numbers is 100.
x-smaller
y-larger
xy=100
y=(100)/(x)
We write a function that represents the minimum sum of two numbers.
f(x,y)=x+y
Substitute y=100+x into f(x,y)=x+y.
f(x)=x+(100)/(x)
=(100x+100)/(x)
Now, we find the first derivative of the function f(x)=x+(100)/(x).
f'(x)=(x+(100)/(x))'
=(x)'+((100)/(x))'
=1-(100)/(x^(2))
Step 2
Find the critical points. Solve the equation f'(x)=0.
f'(x)=1-(100)/(x^(2))
0=1-(100)/(x^(2))
-1=-(100)/(x^(2))
x^(2)=100
x=sqrt(100)=10
x=-sqrt(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-(100)/(x^(2)))'
=(1)'-((100)/(x^(2)))'
=(200)/(x^(3))
Calculate f""(10).
f""(10)=(200)/(1000)=0.2>0
Since, it is positive, it means that yes, there is a minimum.
Calculate y. Substitute x=10 into xy=100.
10y=100
y=10
Best Answer
dromero420 Staff answered 4 weeks ago
Let the numbers be x,y xy=100, y=(100)/(x) f(x,y)=x+y f(x)=x+(100)/(x) Minimum of f(x) is obtained at f'(x)=0 -> f'(x)=1-(100)/(x^(2))=0 x=pm 10 therefore x=10, y=10
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