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).
Is the annual dividend 1 dollar per share? Because if so, that means you would get 500 dollars yearly, and then divide that by 4 to get quarterly, making the answer 125 dollars in the quarterly check.
To find percent decrease and increase you must subtract the original value (old value) with the new value. Then divide the value you get by the original value and multiply by 100:
165-158 = 7
7÷ 165 × 100 = 4.2424....% or 4.2% rounded to 2 significant figures.