The minimum distance is the perpendicular distance. So establish the distance from the origin to the line using the distance formula. The distance here is: <span><span>d2</span>=(x−0<span>)^2</span>+(y−0<span>)^2 </span> =<span>x^2</span>+<span>y^2 </span></span> To minimize this function d^2 subject to the constraint, <span>2x+y−10=0 </span>If we substitute, the y-values the distance function can take will be related to the x-values by the line:<span>y=10−2x </span>You can substitute this in for y in the distance function and take the derivative: <span>d=sqrt [<span><span><span>x2</span>+(10−2x<span>)^2] </span></span></span></span> d′=1/2 (5x2−40x+100)^(−1/2) (10x−40)<span> </span>Setting the derivative to zero to find optimal x, <span><span>d′</span>=0→10x−40=0→x=4 </span> This will be the x-value on the line such that the distance between the origin and line will be EITHER a maximum or minimum (technically, it should be checked afterward). For x = 4, the corresponding y-value is found from the equation of the line (since we need the corresponding y-value on the line for this x-value).
Then y = 10 - 2(4) = 2.
So the point, P, is (4,2).
Due to the high number of deaths and the subsequent population reduction, the merchants - and tradespeople in general - saw a decrease in the sales of their merchandise. Once the number of infected people by the plague reduced was reduced at the end of the Fifteenth Century, the population grew exponentially, thus creating a higher demand for services and goods.
