Find the extreme values of f on the region described by the inequality. f(x, y) = e−xy; x2 + 4y2 ≤ 1

f(x, y) = e

−xy. For the interior of the region, we find the critical points:

fx = −ye−xy

, fy = −xe−xy, so the only critical point is (0, 0), and f(0, 0) = 1.

For the boundary, we use Lagrange multipliers.

g(x, y) = x

2 + 4y

2 = 1 ⇒ λ∇g = h2λx, 8λyi, so setting ∇f = λ∇g we get

−ye−xy = 2λx and −xe−xy = 8λy. The first of these gives e

−xy = −2λx/y,

and then the second gives −x(−2λx/y) = 8λy ⇒ x

2 = 4y

2

. Solving this

last equation with the constraint x

2 + 4y

2 = 1 gives x = ± √

1

2

and y = ±

1

2

√

2

.

Now f

± √

1

2

, ∓

1

2

√

2

= e

1/4 ≈ 1.284 and f

± √

1

2

, ±

1

2

√

2

= e

−1/4 ≈ 0.779.

The former are the maxima on the region and the latter are the minima.

Step-by-step explanation:

step-by-step explanation: yes

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)