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Let's consider an arbitrary point p(x,y)The distance from P to the origin is:d=x2+y2we notice that this the objective function. since p is on the line y=3x+4 the coordinates of P satisties this equation. This is the constrain for the objective function in (1).Therefore, the objective function can be written as a function of one variable:d=d(x)=x2+(3x+4)2From the first devirative tes, we have thatd′(x)=0⇒d2d(x)⋅ddx(x2+(3x+4)2)=0⇒2x+2(3x+4)(3)=0⇒2x+18x+24=0⇒20x=−24⇒x=−2420=−65Since d′(x)<0 for x<−65 and d′(x)>0 for x≻65, the minimum distance from p to the origin is achieved when x=−65Then, y=3x+4=3(−65)+y⇒y=−18+205=25we found that, P(−65,25) is the point on the line y=3x+4 closest to the origin.
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Find the point on the plane x+2y+3z=13 that is closest to the point (1,1,1). How would you minimize the function?
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