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).
Answer:
18x^4 − 12x^3 + 57x^2 − 28x + 35
Answer: You are usually multiplying
Step-by-step explanation: The graph usually curves and not straight.
Attention:THIS IS PART B <em>ONLY!</em>
First,find the unit rate.
35/2 and <em>x</em>/1
35 divided by 2 = 17.5 so the unit rate for J.K. is 17.5 pages per hour.
now,R.L.
45/3 and <em>x</em>/1
45 divided by 3 = 15 so the unit rate for R.L. is 15 pages per hour
now we need to divide 355 by 17.5 and divide it by 15.Whichever quotient is smaller,finishes the book first
355 divided by 17.5 = about 20 hrs----J.K.
355 divided by 15 = about 24 hrs-----R.L.
So,who read faster?
