Please check this for me:

the lenght l, width w, & height h of a box change with time. At a certain instant the dimensions are l=1m & w=h=2m. L and W are increasing at a rate of 2m/sec while H is decreasing at a rate of 2m/sec. At that instant, find the rate at which the surface area is changing.

Heres what I got:
We want the derivative of surface area with respect to t (ds/dt)...
ds/dt= 2y*dx/dt + 2x*dy/dt + 2z*dy/dt+ 2y*dz/dt + 2x*dz/dt + 2z*dx/dt
=2(2m)(-3) + 2(1m)(2) + 2(2m)(2)+ 2(2m)(2)+ 2(1m)(2)+2(2m)(-3)
=-12m+4m+8m+8m+4m+12m
=-24m+24m
=O m^2/sec

Is this right??

To find the rate at which the surface area is changing, we need to use the formula for the surface area of a box:

Surface Area (S) = 2lw + 2lh + 2wh

We are given the values of l, w, and h at a certain instant as l=1m, w=h=2m. In this case, the surface area is:

S = 2(1)(2) + 2(1)(2) + 2(2)(2) = 4 + 4 + 8 = 16

Next, we have the rates of change for l, w, and h. L and W are increasing at a rate of 2m/sec, while H is decreasing at a rate of 2m/sec.

To find the rate at which the surface area is changing, we can differentiate the formula for surface area with respect to time (t):

dS/dt = 2(dl/dt)w + 2l(dw/dt) + 2(dh/dt)w + 2h(dw/dt) + 2l(dh/dt) + 2h(dl/dt)

Now, substitute the given values and rates of change:

dS/dt = 2(2)(2) + 2(1)(2) + 2(-2)(2) + 2(2)(2) + 2(1)(-2) + 2(2)(2)
= 8 + 4 - 8 + 8 - 2 + 8
= 18

Therefore, the rate at which the surface area is changing at that instant is 18 square meters per second.