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Sloan [31]
3 years ago
5

Find two positive numbers whose product is 81 and whose sum is a minimum.

Mathematics
1 answer:
hammer [34]3 years ago
3 0
Ab=81, a=81/b

s=a+b using a from above in this we get:

s(b)=81/b +b

s(b)=(81+b^2)/b

ds/db=(2b*b-81-b^2)/b^2

ds/db=(2b^2-81-b^2)/b^2

ds/db=(b^2-81)/b^2

d2s/db2=(2b^3-2b^3+81)/b^4

d2s/db2=81/b^4 since b is positive we know that the acceleration is positive so that when ds/db=0 it is a minimum for s(b)

ds/db=0 only when b^2-81=0, b^2=81, b=9

The two positive numbers are 9 and 9.


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