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Divide 20 into two parts such that the sum of the parts of the squares is 208. |
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Answer» x+y=20 and z=xy3 ⇒z=y3(20-y)=20y3-y4 ⇒ dz dy
=60y2-4y3=0⇒4y2(15-y)=0 So, either y=0,ory=15 Now d2z dy2
=120y-12y2,bacauseAty=0, d2z dy2
>0 bacausey=0 is the POINT of minima and at y=15, d2z dy2
<0 ∵y=15 is the point of maximum. Hence the REQUIRED parts is (5,15) Step-by-step EXPLANATION: x+y=20 and z=xy3 ⇒z=y3(20-y)=20y3-y4 ⇒ dz dy
=60y2-4y3=0⇒4y2(15-y)=0 So, either y=0,ory=15 Now d2z dy2
=120y-12y2,bacauseAty=0, d2z dy2
>0 bacausey=0 is the point of minima and at y=15, d2z dy2
<0 ∵y=15 is the point of maximum. Hence the required parts is (5,15) |
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