Given that f(x, y) is a linear function, and the constraints themselves are also linear, it follows that the vertices of the feasible region are the sites of extrema of f(x, y). So just find where each boundary line intersects with another line, and check the value of f(x, y) at each intersection.

We have

• x = 0 and y = 0 ===> (0, 0)

• x = 0 and 2x + 7y = 70 ===> y = 10 ===> (0, 10)

• y = 0 and 8x + 4y = 136 ===> x = 17 ===> (17, 0)

• 2x + 7y = 70 and 8x + 4y = 136 ===> (14, 6)

At these points, we respectively get

• f (0, 0) = 0

• f (0, 10) = 60

• f (17, 0) = 34

• f (14, 6) = 64

Then max f(x) = 64 at (14, 6) and min f(x) = 0 at (0, 0).

5 0

3 years ago

Which of the following is a solution of x2 + 2x + 8?

____ [38]

A solution of a quadratic equation (ax^2+bx+c) is the point at which the parabola crosses the x-axis. We can find this by using the Quadratic formula, which is $\frac{-b+- \sqrt{b^2-4ac} }{2a}$ . We can solve the equation as follows:

$\frac{-b+- \sqrt{b^2-4ac} }{2a} \\ \frac{-2+- \sqrt{2^2-4(1)(8)} }{2(1)} \\ \frac{-2+- \sqrt{4-32} }{2} \\ \frac{-2+- \sqrt{-28} }{2}$
Then we separate the negative from -28 to get:
$\frac{-2+- \sqrt{28}* \sqrt{-1} }{2}=\frac{-2+-2i \sqrt{7} }{2}$
Then we continue to solve by factoring common terms (-2 and 2). We get the solutions of $-1+i \sqrt{7} \\ or \\ -1-i \sqrt{7}$ . Choice B matches our first solution.

:)

7 0

3 years ago