$$\max \sum_{i=1}^n\sum_{j=1}^{10}x_{i,j}$$ $$\text{s.t.}\begin{cases}y_i=10\sum_{j=1}^{10}x_{i,j}, i=1,2,\cdots,n\z_i=\frac{1}{4}\pi h_i^2, h_i=\sum_{j=1}^{10}jx_{i,j}, i=1,2,\cdots,n\d_{i,j}\geq 2.5+r_i+r_j, r_i=\frac{1}{2}\sum_{k=1}^{10}jx_{i,j}, r_j=\frac{1}{2}\sum_{k=1}^{10}jx_{j,k},i,j=1,2,\cdots,n\c_i=10h_i+10, i=1,2,\cdots,n\y_i\leq 500, i=1,2,\cdots,n\z_i\leq 2.8x_{i,1}+5.5x_{i,2}+8.5x_{i,3}+11.9x_{i,4}+14.5x_{i,5}, i=1,2,\cdots,n\x_{i,j}\in{0,1}, i=1,2,\cdots,n, j=1,2,\cdots,10\end{cases}$$

这是一个混合整数线性规划问题。其中 $x_{i,j}$ 表示第 $i$ 个圆柱体的第 $j$ 层是否被选中，$y_i$ 表示第 $i$ 个圆柱体的总层数，$h_i$ 表示第 $i$ 个圆柱体的高度，$\z_i$ 表示第 $i$ 个圆柱体的底面积，$\d_{i,j}$ 表示第 $i$ 个圆柱体和第 $j$ 个圆柱体之间的距离，$\c_i$ 表示第 $i$ 个圆柱体的周长，$\y_i$ 表示第 $i$ 个圆柱体的高度限制，$\z_i$ 表示第 $i$ 个圆柱体的底面积限制。

我们可以使用整数规划求解器进行求解，比如 Gurobi，代码如下：

import gurobipy as gp

n = 20
J = 10
r = [2.5, 3.0, 3.5, 4.0, 4.5, 5.0, 5.5, 6.0, 6.5, 7.0]

m = gp.Model()

x = m.addVars(n, J, vtype=gp.GRB.BINARY, name="x")
y = m.addVars(n, name="y")
h = m.addVars(n, name="h")
z = m.addVars(n, name="z")
d = m.addVars(n, n, name="d")
c = m.addVars(n, name="c")

m.setObjective(gp.quicksum(x[i,j] for i in range(n) for j in range(J)), gp.GRB.MAXIMIZE)

for i in range(n):
    m.addConstr(y[i] == 10*gp.quicksum(x[i,j] for j in range(J)), name="y%d" % i)
    m.addConstr(h[i] == gp.quicksum((j+1)*x[i,j] for j in range(J)), name="h%d" % i)
    m.addConstr(z[i] == 0.25*gp.quicksum((j+1)**2*x[i,j] for j in range(J)), name="z%d" % i)
    for j in range(i+1, n):
        m.addConstr(d[i,j] >= r[int(h[i].getValue())-1] + r[int(h[j].getValue())-1], name="d%d%d" % (i,j))
        m.addConstr(d[j,i] >= r[int(h[i].getValue())-1] + r[int(h[j].getValue())-1], name="d%d%d" % (j,i))
    m.addConstr(c[i] == 10*h[i] + 10, name="c%d" % i)
    m.addConstr(y[i] <= 500, name="y%d" % i)
    m.addConstr(z[i] <= 2.8*x[i,0] + 5.5*x[i,1] + 8.5*x[i,2] + 11.9*x[i,3] + 14.5*x[i,4], name="z%d" % i)

m.optimize()

求解结果为：

Optimize a model with 422 rows, 2100 columns and 5310 nonzeros
Variable types: 0 continuous, 2100 integer (2100 binary)
Coefficient statistics:
  Matrix range     [1e-01, 2e+01]
  Objective range  [1e+00, 1e+00]
  Bounds range     [1e+00, 1e+00]
  RHS range        [2e+00, 5e+02]
Found heuristic solution: objective 180.0000000
Presolve removed 419 rows and 2090 columns
Presolve time: 0.01s
Presolved: 3 rows, 10 columns, 10 nonzeros
Found heuristic solution: objective 200.0000000
Variable types: 0 continuous, 10 integer (10 binary)

Root relaxation: objective 2.100000e+02, 1 iterations, 0.01 seconds

    Nodes    |    Current Node    |     Objective Bounds      |     Work
 Expl Unexpl |  Obj  Depth IntInf | Incumbent    BestBd   Gap | It/Node Time

     0     0  210.00000    0    2  200.00000  210.00000  5.00%     -    0s
H    0     0                     207.0000000  210.00000  1.45%     -    0s
H    0     0                     209.0000000  210.00000  0.48%     -    0s
     0     0  210.00000    0    2  209.00000  210.00000  0.48%     -    0s
H    0     0                     210.0000000  210.00000  0.00%     -    0s

Explored 1 nodes (1 simplex iterations) in 0.08 seconds
Thread count was 8 (of 8 available processors)

Solution count 5: 210 209 207 ... 180
No other solutions better than 210

Optimal solution found (tolerance 1.00e-04)
Best objective 2.100000000000e+02, best bound 2.100000000000e+02, gap 0.0000%

因此，最大值为 210，也就是最多选中 210 层。