2x+2y=20 −> y=10−xS=xy=x(10−x)dSdx=10−2xdSdx=0 −> x=5, y=10−x=5.Point (5,5) is the maximum point because d2Sdx2=−2<02x+2y=20\:\;->\;\;y=10-x\\ S=xy=x(10-x)\\ \frac{dS}{dx}=10-2x\\ \frac{dS}{dx}=0\;\;->\;\;x=5,\;\;y=10-x=5.\\ Point\;(5,5)\;is\ the\ maximum\ point\ because\;\;\frac{d^2S}{dx^2}=-2<02x+2y=20−>y=10−xS=xy=x(10−x)dxdS=10−2xdxdS=0−>x=5,y=10−x=5.Point(5,5)is the maximum point becausedx2d2S=−2<0
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