Answer:7x + -5y = -28
(-4, _)
(1, _)
(_, 14)
Step-by-step explanation:
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
.
Drawing this square and then drawing in the four radii from the center of the cirble to each of the vertices of the square results in the construction of four triangular areas whose hypotenuse is 3 sqrt(2). Draw this to verify this statement. Note that the height of each such triangular area is (3 sqrt(2))/2.
So now we have the base and height of one of the triangular sections.
The area of a triangle is A = (1/2) (base) (height). Subst. the values discussed above, A = (1/2) (3 sqrt(2) ) (3/2) sqrt(2). Show that this boils down to A = 9/2.
You could also use the fact that the area of a square is (length of one side)^2, and then take (1/4) of this area to obtain the area of ONE triangular section. Doing the problem this way, we get (1/4) (3 sqrt(2) )^2. Thus,
A = (1/4) (9 * 2) = (9/2). Same answer as before.
Consecutive angles are supplementary (A + D = 180°). If one angle is right, then all angles are right. The diagonals of a parallelogram<span> bisect each other. Each </span>diagonal of a parallelogram separates it into two congruent triangles.<span> A square has got 4 sides of equal length and 4 right </span>angles<span> (right </span>angle<span> = 90 degrees). Since ALL the </span>angles<span> in a</span>quadrilateral add up<span> to 360 then 360 divided by 4 must be 90.</span>
18 - 22 + 5x(66:3)
= 18 - 22 + 5x22
= -4 + 110
= 77
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