I assume there are some plus signs that aren't rendering for some reason, so that the plane should be

.
You're minimizing

subject to the constraint

. Note that

and

attain their extrema at the same values of

, so we'll be working with the squared distance to avoid working out some slightly more complicated partial derivatives later.
The Lagrangian is

Take your partial derivatives and set them equal to 0:

Adding the first three equations together yields

and plugging this into the first three equations, you find a critical point at

.
The squared distance is then

, which means the shortest distance must be

.
∠BDC and ∠AED are right angles, is a piece of additional information is appropriate to prove △ CEA ~ △ CDB
Triangle AEC is shown. Line segment B, D is drawn near point C to form triangle BDC.
<h3> What are Similar triangles?</h3>
Similar triangles, are those triangles which have similar properties,i.e. angles and proportionality of sides.
Image is attached below,
as shown in figure
∡ACE = ∡BCD ( common angle )
∡AED = ∡BDC ( since AE and BD are perpendicular to same line EC and make right angles as E and C)
∡EAC =- ∡DBC ( corresponding angles because AE and BD are parallel lines)
Thus, △CEA ~ △CDB , because of the two perpendiculars AE and BD.
Learn more about similar triangles here:
brainly.com/question/25882965
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Answer:
18/2 : 9
Step-by-step explanation:
As you probably know there is a difference between mean and median -- the mean acts as a measure of the average, and on the other hand median measures the middle number. The median in this is 15 and the mean is also 15.