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:
See explanation
Step-by-step explanation:
Simplify left and right parts separately.
<u>Left part:</u>
<u>Right part:</u>
Since simplified left and right parts are the same, then the equality is true.
Answer and Step-by-step explanation:
Each package has 12 pencils
total number of pencils Sarah purchased: n = 12p
n = total number of pencils
p = number of packages
n depends on p, because the total number of pencils dependos on how many packages she will buy, so:
independent variable: p
dependent variable: n
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║ p ║ n = 12p ║ n ║
╠═══╬══════════╬════╣
║ 3 ║ n = 12*3 ║ 36 ║
╠═══╬══════════╬════╣
║ 4 ║ n = 12*4 ║ 48 ║
╠═══╬══════════╬════╣
║ 5 ║ n = 12*5 ║ 60 ║
╠═══╬══════════╬════╣
║ 6 ║ n = 12*6 ║ 72 ║
╠═══╬══════════╬════╣
║ 7 ║ n = 12*7 ║ 84 ║
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