Need answers for A and B, also please explain, I don't understand We'll be analyzing the surface area of a round cylinder - in other words the amount of material needed to"make a can".A cylinder (round can) has a circular base and a circular top with vertical sides in between. Let r be theradius of the top of the can and let h be the height. The surface area of the cylinder, A, isA = 2rr2 .+ 2rrh (it's two circles for the top and bottom plus a rolled up rectangle for the side).r = radiusAreas = T r?h = heightArea = h(2tr)Circumference2arPart a: Assume that the height of your cylinder is 8 inches. Consider A as a function of r, so we canwrite thatA (r) = 2 Tr- + 16 r. What is the domain of A (r)? In other words, for which values ofr is A (r) defined?Part b: Find the inverse function to A(r). Your answer should look like r ="some expression involvingА".r (A) =Hints:• To calculate an inverse function, you need to solve for r.with A = 2 Tr2This equation isa2?• Hereyoucouldstart+ 16 Tr.thesameas2 Tr + 16 Tr – A = 0. Do you recognize this as a quadratic equation ax + bx +c = 0 wherethe variable x is r? The coefficients would be 2 T for a, 16 T for b, and -A for c.• You can solve for r using the quadratic formula even though the constant term c is a symbol here.

Name: Need answers for A and B, also please explain, I don't understand We'l
Uploaded: 2024-03-20T07:07:12.000Z
Channel: Bartleby
Description: Need answers for A and B, also please explain, I don't understand We'll be analyzing the surface area of a round cylinder - in other words the amount of material needed to"make a can".A cylinder (round can) has a circular base and a circular top with

Part a) Ar=2πr2+16πr As area cannot be negative or zero. So, Ar>0 i.e., 2πrr+8>0 ⇒r>0 or r>-8Thus, domain= r>0 or r∈0,∞ Part b) Here, A(r)=2πr2+16πri.e., 2πr2+16πr-A=0 Comparing this equation with ax2+bx+c=0where, a=2π , b=16π and c=-ASolving for r using completing squares, r2+16πr2π-A2π=0r2+8r-A2π=0r2+8r+42-42-A2π=0r+42-16-A2π=0 r+42=16+A2π r+4=±16+A2π r =±32π+A2π-4 Thus , inverse function to A(r)= ±32π+A2π-4