# Tag Info

1

I haven't checked if your solution is the right solution for the problem, but this is the linear programming equivalent of your program: LinearProgramming[-{0.0509, 0.078, 0.06875, 0.0864, 0.026}, {{-.4, .6, .6, -.4, -.4}, {-1, 0, 0, 1, -1}, {-.03, .02, -.01, .01, -.06}, {-1, -1, -1, -1, -1}}, ...

3

funcs = Tooltip /@ ({E^x, (Series[E^x, {x, 0, #}] // Normal) & /@ Range[6]} // Flatten); Plot[funcs, {x, 0, 5}]

-1

Sorry, I somehow have two accounts one on my computer the other my phone. I can't edit or comment, so I reply with this update for you: Actually, I am supposed to use cylindrical coordinates if needed. Also could you not use Integrate[ h(x,y,z,),(x,x_o,x_f},{y,..... Hand typed so it isn't exact.

5

By adding axes labels to your plot we can see that the base of the cone is not in the zy plane as requested. You need to rework your inequality so that the radius is proportional to the distance from origo along the x-axis, i.e. $z^2+y^2\leq x^2$. RegionPlot3D[ (y^2 + z^2)^(1/2) <= x && 0 <= x <= 5, {x, 0, 5}, {y, -5, 5}, {z, -5, 5}, ...

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