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).
For this case, the first thing we must take into account is the following definition:
d = v * t
Where,
d: distance
v: speed
t: time
Substituting values we have:
Trip 1:
50 = v * t1
We cleared t1:
t2 = (50) * (1 / v)
Trip 2:
300 = (3 * v) * t2
We cleared t2:
t2 = (300/3) * (1 / v)
t2 = (100) * (1 / v)
Rewriting:
t2 = 2 (50) * (1 / v)
t2 = 2 * t1
Answer:
His new time compared with the old time was:
Twice the old time.
t2 = 2 * t1.
x + 4 = 12 Subtract 4 on both sides to get "x" by itself
x + 4 - 4 = 12 - 4
x = 8
Answer:
Step-by-step explanation:
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