Let ii be the tangent plane to the graph of f(x, y) = 8 – 2x^2 – 3y^2 at the point (1, 2,-6). let s, x² + y^2 + z = 4 be another surface. find the point on s which has tangent plane parallel to i.

Let . The tangent plane to at (1, 2, -6) has equation

We have

Then the tangent plane has equation

Let , and . The tangent plane to at a point is

We have

so that this plane has equation

In order for this plane to be parallel to the previous plane, we need to have

so the point we're looking for is (2, 6, -36).

step-by-step explanation:

given perimeter: 28cm

length: 9cm

atp,

2(l+w)=p of the reactangle

2(9+x)=28

18+2x=28

2x=28-18

x=10/2

x=5

therefore width is 5 cm