Question

Question #36079

Answer 1

If `y` equals `x`, let's substitute `x` for `y` so we only have `2` variables instead of three

`z = 3x +x -x^2`

`z = x^2+4x`

or

`z = y^2 + 4y`

graph{y = x^2 + 4x}

So, since this quadratic is increasing, that means the highest point is at the greatest endpoint (`y=4`)

So let's solve for that first:

`z = (4)^2 + 4(4)`

`z = 16+16 = 32`

Now let's solve for when `x=0`

`z = 0^2 + 4(0)`

`z = 0`

So the maximum value is `z = 32` and the minimum is `z = 0`