How do you find two positive numbers whose product is 750 and for which the sum of one and 10 times the other is a minimum?

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Answer 1 · 2018-07-12T21:33:24

#50\sqrt3# & #5\sqrt3#

Let the positive numbers be #x# & #y# such that

#xy=750\implies y=750/x#

Let #S# be the sum of #x# & #10# times #y# then we have

#S=x+10y#

#S=x+10(750/x)#

#S=x+7500/x#

#\frac{d}{dx}S=\frac{d}{dx}(x+7500/x)#

#\frac{dS}{dx}=1-7500/x^2#

#\frac{d^2S}{dx^2}=15000/x^3#

for minimum value of #S# we have #\frac{dS}{dx}=0# as follows

#1-7500/x^2=0#

#x=\pm50\sqrt3#

But #x, y>0# therefore we have #x=50\sqrt3#. Now, we have

#(\frac{d^2S}{dx^2})_{x=50\sqrt3}=15000/(50\sqrt3)^3>0#

hence, the sum #S# is minimum at #x=50\sqrt3#

#\implies y=750/x#

#=750/{50\sqrt3}#

#=5\sqrt3#

Hence, the positive numbers are #50\sqrt3# & #5\sqrt3#