Solve for q: (4q-6n)/(3m)=f
1 answer:
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The objective function is simply a function that is meant to be maximized. Because this function is multivariable, we know that with the applied constraints, the value that maximizes this function must be on the boundary of the domain described by these constraints. If you view the attached image, the grey section highlighted section is the area on the domain of the function which meets all defined constraints. (It is all of the inequalities plotted over one another). Your job would thus be to determine which value on the boundary maximizes the value of the objective function. In this case, since any contribution from y reduces the value of the objective function, you will want to make this value as low as possible, and make x as high as possible. Within the boundaries of the constraints, this thus maximizes the function at x = 5, y = 0.
Answer:
2x - y - 3z = 0
Step-by-step explanation:
Since the set
{i, j} = {(1,0), (0,1)}
is a base in
and F is linear, then
<em>{F(1,0), F(0,1)} </em>
would be a base of the plane generated by F.
F(1,0) = a+b = (2i-2j+k)+(i+2j+k) = 3i+2k
F(0,1) = a+c = (2i-2j+k)+(2i+j+2k) = 4i-j+3k
Now, we just have to find the equation of the plane that contains the vectors 3i+2k and 4i-j+3k
We need a normal vector which is the cross product of 3i+2k and 4i-j+3k
(3i+2k)X(4i-j+3k) = 2i-j-3k
The equation of the plane whose normal vector is 2i-j-3k and contains the point (3,0,2) (the end of the vector F(1,0)) is given by
2(x-3) -1(y-0) -3(z-2) = 0
or what is the same
2x - y - 3z = 0