Maximizing the volume of a box ============================== *This example is adapted from Boyd, Kim, Vandenberghe, and Hassibi,* "`A Tutorial on Geometric Programming `__\ ". In this example, we maximize the shape of a box with height :math:`h`, width :math:`w`, and depth :math:`d`, with limits on the wall area :math:`2(hw + hd)` and the floor area :math:`wd`, subject to bounds on the aspect ratios :math:`h/w` and :math:`w/d`. The optimization problem is .. math:: \begin{array}{ll} \mbox{maximize} & hwd \\ \mbox{subject to} & 2(hw + hd) \leq A_{\text wall}, \\ & wd \leq A_{\text flr}, \\ & \alpha \leq h/w \leq \beta, \\ & \gamma \leq d/w \leq \delta. \end{array} .. code:: python import cvxpy as cp # Problem data. A_wall = 100 A_flr = 10 alpha = 0.5 beta = 2 gamma = 0.5 delta = 2 h = cp.Variable(pos=True, name="h") w = cp.Variable(pos=True, name="w") d = cp.Variable(pos=True, name="d") volume = h * w * d wall_area = 2 * (h * w + h * d) flr_area = w * d hw_ratio = h/w dw_ratio = d/w constraints = [ wall_area <= A_wall, flr_area <= A_flr, hw_ratio >= alpha, hw_ratio <= beta, dw_ratio >= gamma, dw_ratio <= delta ] problem = cp.Problem(cp.Maximize(volume), constraints) print(problem) .. parsed-literal:: maximize h * w * d subject to 2.0 * (h * w + h * d) <= 100.0 w * d <= 10.0 0.5 <= h / w h / w <= 2.0 0.5 <= d / w d / w <= 2.0 .. code:: python assert not problem.is_dcp() assert problem.is_dgp() problem.solve(gp=True) problem.value .. parsed-literal:: 77.45966630736292 .. code:: python h.value .. parsed-literal:: 7.7459666715289766 .. code:: python w.value .. parsed-literal:: 3.872983364643079 .. code:: python d.value .. parsed-literal:: 2.581988871583608 .. code:: python # A 1% increase in allowed wall space should yield approximately # a 0.83% increase in maximum value. constraints[0].dual_value .. parsed-literal:: 0.8333333206334043 .. code:: python # A 1% increase in allowed wall space should yield approximately # a 0.66% increase in maximum value. constraints[1].dual_value .. parsed-literal:: 0.6666666801983365