Asilo (base not included ) is to be constructed in the form of a cylinder surmounted by a hemisphere. the cost of construction per square unit of surface area is 5 times as great fit the hemisphere as it is for the cylindrical sidewall. determine the dimensions to be used if the volume is fixed at 8,000 cubic units and the cost of construction is to be kept at minimum. neglect the thickness of the silo and waste in construction. i have the function of the cost as c=(16000/r) + (26pi r^2 /3) it is now asking for the interval of interest. .in interval notation. i can see that it can't be zero so i have ( , but i have no idea how to find end point on the interval of interest.!

I've worked the problem and I also get the cost equation of

c(r) = 16000/r + 26/3*pi*r^2

which is basically identical to what you've gotten. So I'm not going to bother explaining how to get to this point.

But since you're looking for a minimum, that just screams FIRST DERIVATIVE. And for a simple function such as this, that's easy to get. So let's do it.
c(r) = 16000/r + 26/3*pi*r^2

Rewrite first term to make power explicit, so
c(r) = 16000*r^(-1) + 26/3*pi*r^2

Now for simple functions such as this, just multiply each term's coefficient by the exponent, then subtract 1 from the exponent. So we get
c'(r) = -16000*r^(-2) + 52/3*pi*r
c'(r) = -16000/r^2 + 52/3*pi*r

And that's the first derivative of your cost function. Your desired radius will be the value r where the value of the first derivative is 0. So
c'(r) = -16000/r^2 + 52/3*pi*r
0 = -16000/r^2 + 52/3*pi*r
16000/r^2 = 52/3*pi*r
16000=52/3*pi*r^3
48000/(52*pi) = r^3
12000/(13*pi) = r^3
cuberoot(12000/(13*pi)) = r
r is approximately 6.648076481

X=3 and y=4 is the answer