Python3, 94 bytes

lambda a,b,h:sum(min((L:=abs(A-h[-(i+1)]))*a,L*b)for i,A in enumerate(h[:len(h)//2+len(h)%2]))

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Python3, 246 bytes

Brute-force version with some optimizations:

def f(a,b,h):
 q,s,S=[(h,0)],-1,[]
 for h,c in q:
  if h==h[::-1]:s=c;continue
  for x in range(len(h)):
   for C,O in[(a,1),(b,-1)]:
    H=[*h];H[x]+=O
    if all(a!=H or C+c<u for a,u in S)and(s==-1 or C+c<s):S+=[(H,c+C)];q+=[(H,c+C)]
 return s

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JavaScript (ES6), 70 bytes

f=(a,g,s)=>a.reduce((t,h,i)=>t+Math.abs(a[a.length+~i]-h),0)*(g>s?s:g)/2

Double-counts the differences, but we only need to make half the changes, which is achieved by dividing by 2 at the end.

05AB1E, 20 bytes

WU|vy#‚ø€¥˜Ä2ä`OX*,

Explanation

WU                    # store minimum cost in X
  |v                  # for each testcase
    y#                # convert to list split on spaces
      ‚ø             # zip with its own reverse
         €¥           # deltas of each pair
           ˜Ä         # deep flatten and take absolute value
             2ä       # split into 2 pieces
               `      # flatten (leaves the smaller piece on top of the stack)
                O     # sum
                 X*   # multiply with minimum cost
                   ,  # print

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Julia, 88 Bytes

f(l,A,B)=reduce(+,map(abs,map(sum,[zip(-l,reverse(l))...][1:fld(end,2)])))*min([A,B]...)

Explained:

If I am not mistaken, all we have to do is either grow all necessary plants or shrink them (depending on which operations is less expensive)

1. zip input list with a reverse of itself (align corresponding heights)
2. only look at first half of the list (floor when length is odd)
3. sum the differences
4. multiply with minimum cost (A or B)

Example

f([1,6,2,3,4],1,2) -> 6   (grow plant 1 and 4, yields [4,6,2,6,4])
f([1,6,2,3,4],3,2) -> 12   (shrink plant 2 and 5, yields [1,3,2,3,1])

Note:

As there is still a discussion about the input format, I disregarded it for the moment.

Update:

Updated to output energy, as I think that outputing the actual palindrome is more challenging I attach my previous code here (76 Bytes):

f(l,A,B)=reduce(hcat,map(x->sort([x...]),[zip(l,reverse(l))...]))[A>B?1:2,:]