Question #192473

find two positive numbers whose product is 64 and whose sum is a minimum


1
Expert's answer
2021-05-13T11:55:50-0400

Let the numbers be x and y


According to question-


xy=64y=64x       (1)xy=64\Rightarrow y =\dfrac{64}{x}~~~~~~~-(1)


Also Let S be their sum-


S=x+yS=x+64xS=x+y \\[9pt] S=x+\dfrac{64}{x}


Differentiate w.r.t. x-


dSdx=164x2\dfrac{dS}{dx}=1-\dfrac{64}{x^2}


Putting dSdx=0164x2=0x2=64x=±8\dfrac{dS}{dx}=0\Rightarrow 1-\dfrac{64}{x^2}=0\Rightarrow x^2=64\Rightarrow x=\pm 8


Again differentiating dSdx\dfrac{dS}{dx} w.r.t. xx-


d2Sdx2=128x3\dfrac{d^2S}{dx^2}=\dfrac{128}{x^3}


d2Sdx2 at x=8=12883=2\dfrac{d^2S}{dx^2}_{\text{ at } x=8}=\dfrac{128}{8^3}=2


So d2Sdx2>0\dfrac{d^2S}{dx^2}>0


SO, S is miniumum at x=8


Then Other number y=648=8y=\dfrac{64}{8}=8


Hence The two numbers are 8 and 8.

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