We are asked to find the values of

x

or

y

that minimize the given function. Notice that the third set of constraints,

x \geq 0

and

y \geq 0,

limit the solution set to the first quadrant. This means we should find the solution set of the first and second inequality in the first quadrant. Let's start by isolating

y

in each inequality.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 4 x + 3 y \geq 30 x + 3 y \geq 21 x \geq 0, y \geq 0 (I) (II) (III)

▼

(I), (II):

Write in slope-intercept form

SubEqn

$(I):$ $LHS - 4 x = RHS - 4 x$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 3 y \geq - 4 x + 30 x + 3 y \geq 21 x \geq 0, y \geq 0

DivEqn

$(I):$ $LHS / 3 = RHS / 3$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ y \geq - \frac{4 x}{3} + 10 x + 3 y \geq 21 x \geq 0, y \geq 0

SubEqn

$(II):$ $LHS - x = RHS - x$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ y \geq - \frac{4 x}{3} + 10 3 y \geq - x + 21 x \geq 0, y \geq 0

DivEqn

$(II):$ $LHS / 3 = RHS / 3$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ y \geq - \frac{4 x}{3} + 10 y \geq - \frac{x}{3} + 7 x \geq 0, y \geq 0

MovePartNumRight

$(II):$ $\frac{a \cdot b}{c} = \frac{a}{c} \cdot b$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ y \geq - \frac{4}{3} x + 10 y \geq - \frac{1}{3} x + 7 x \geq 0, y \geq 0

To solve the system of inequalities we need to enter them in the calculator. Push

Y =

and write the functions in the first two rows.

Next, we have to set the calculator to account for the inequality. Place the cursor before $Y$ and scroll by pushing $ENTER .$ We will select the icon that displays a black triangle pointing up and to the right. This tells the calculator that we are dealing with a $\geq$ inequality.

With the inequalities entered in the calculator, we can graph them by pushing $GRAPH .$ Before we do that let's limit the window to the first quadrant, as we have already established that we are interested in this area only.

The objective function is minimized where the boundary lines of the two first inequalities intersect. We can use the calculator to find this point of intersection. To do this, push CALC ( $2nd + TRACE$ ) and choose the fifth option in the list, intersect.

Now we must provide the calculator with some guesses as to where the intersection occurs.

The solution is $x = 3$ and $y = 6 .$ These values minimize the objective function.

Exercise