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:
13
Step-by-step explanation:
Diagonals of a rhombus are perpendicular bisector.
Hence, triangle ABE so formed would be a right triangle right angled at E.
Therefore, by Pythagoras theorem:
I'm assuming that the 1/3 is an exponent.
If so, then
![2^{1/3} = \sqrt[3]{2}](https://tex.z-dn.net/?f=2%5E%7B1%2F3%7D%20%3D%20%5Csqrt%5B3%5D%7B2%7D)
Which is the cube root of 2. Raising any value to the 1/3 power is the same as taking the cube root.
Answer:
30 tons
Step-by-step explanation:
