根据@JRose和an answer to a related question的评论，我发现这是一个the Knapsack Problem的例子。这个想法是比一些具有不同价值和大小的对象需要打包到一个背包中，以便在满足大小限制的情况下最大化总价值。
在我的示例中，MyClass的所有示例都具有相同的值，但具有不同的大小（称为面积）。我在SciPy文档中找到了一个工作示例，并添加了一些修改，包括目标区域的软边界。最终，这满足了我的要求。
代码如下所示：

import numpy as np
from scipy import optimize
from scipy.optimize import milp
sizes = np.array([0.003968, 0.0148, 0.022912, 0.024832, 0.02912, 0.032448,
                  0.034176, 0.035136, 0.035584, 0.049344, 0.057984, 0.059904,
                  0.063744, 0.080064, 0.085824])  # tot_area = 0.62984
target_area = 0.45
pct_error = 0.03  # Acceptable error as percentage of target_area.
# All instances of MyClass have equal value.
values = np.full_like(sizes, 1.0)
# Set decision variable (x_i) to be between 0 and 1 and to be integers.
# This makes them effectively Boolean, and they can be cast to bool in the end.
integrality = np.full_like(values, True)  # x_i must be integers within bounds.
bounds = optimize.Bounds(0, 1)  # x_i must be between 0 and 1.
# Calculate and set the upper and lower bounds.
lb = (1 - pct_error) * target_area
ub = (1 + pct_error) * target_area
constraints = optimize.LinearConstraint(A=sizes, lb=lb, ub=ub)
optimization_result = milp(c=-values, constraints=constraints,
                           integrality=integrality, bounds=bounds)
if not optimization_result.success:
    raise RuntimeError('The MILP optimization procedure failed!')
print(f'Total area: {sizes[optimization_result.x.astype(bool)].sum()}' )
#  0.45998399999999995
print(optimization_result.x)
# array([0., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 0., 0.])
print( sizes[optimization_result.x.astype(bool)] )
# [0.0148 0.022912 0.024832 0.02912 0.032448 0.034176 0.035136 0.035584 0.049344 0.057984 0.059904 0.063744]

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