Consider the following problem, which is the Hydit-Filit problem with one of the constraints removed. Max: Z(x)= 140x₁+ 160 x2 s.t. 5x,+ X1 XI 5x2 ≤ 50 ≤8 X2 6 ≥ 0 X2 ≥ 0 I find that the optimal solutions is x* = (4, 6). (U₁) (U₂) (U3) (U4) (us)

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Q: Solve the problem maxx + 4y + z subject to x² + y² + z² = 216 and x + 2y + 3z = 0

A: Dur to insufficient information of part (b) ,we will provide the solution only part(a) .

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A: We will find out the required solution.

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Q: 3. (Sec. 3.5) Consider the nonhomogeneous second order linear DE Ly] =y" +p(t)y' + g(t)y=g(t), where…

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