Question

solve the system
$3x + y = 19$
$x - y = 3$
\bigcirc $(5.5,2.5)$
\bigcirc $(5,3)$
\bigcirc $(4.5,3)$
\bigcirc $(5.5,5.5)$

question 15
1 pts
a company makes tables (t) and chairs (c).
constraints:
$4t + 2c \leq 40$
$t + c \leq 15$
$t \geq 0$
$c \geq 0$
profit equation: $p = 30t + 20c$
choose the combination that maximizes profit.
\bigcirc $t = 8$ $c = 7$
\bigcirc $t = 0$ $c = 15$
\bigcirc $t = 5$ $c = 10$
\bigcirc $t = 10$ $c = 0$

Explanation:

Solving the System of Equations (First Problem)

Step1: Add the two equations

We have the system:

$$\begin{cases} 3x + y = 19 \\ x - y = 3 \end{cases}$$

Adding the two equations to eliminate \(y\):
$$(3x + y)+(x - y)=19 + 3$$
Simplify: \(4x=22\)

Step2: Solve for \(x\)

Divide both sides by 4:
$$x=\frac{22}{4}=5.5$$

Step3: Solve for \(y\)

Substitute \(x = 5.5\) into \(x - y=3\):
$$5.5-y = 3$$
Subtract 5.5 from both sides: \(-y=3 - 5.5=-2.5\)
Multiply both sides by -1: \(y = 2.5\)

We check each combination against the constraints \(4T + 2C\leq40\), \(T + C\leq15\), \(T\geq0\), \(C\geq0\) and calculate the profit \(P = 30T+20C\). The combination \(T = 5\), \(C = 10\) satisfies all constraints and gives the highest profit.

Answer:

(5.5, 2.5)

Maximizing Profit (Second Problem)

We check each option against the constraints and calculate the profit.

Option 1: \(T = 8\), \(C=7\)

• Check \(4T + 2C\leq40\): \(4(8)+2(7)=32 + 14 = 46>40\) (Violates constraint)

Option 2: \(T = 0\), \(C = 15\)

• Check \(4T+2C\leq40\): \(4(0)+2(15)=30\leq40\)
• Check \(T + C\leq15\): \(0 + 15=15\leq15\)
• Profit: \(P=30(0)+20(15)=300\)

Option 3: \(T = 5\), \(C = 10\)

• Check \(4T + 2C\leq40\): \(4(5)+2(10)=20 + 20 = 40\leq40\)
• Check \(T + C\leq15\): \(5+10 = 15\leq15\)
• Profit: \(P=30(5)+20(10)=150 + 200 = 350\)

Option 4: \(T = 10\), \(C = 0\)

• Check \(4T+2C\leq40\): \(4(10)+2(0)=40\leq40\)
• Check \(T + C\leq15\): \(10+0 = 10\leq15\)
• Profit: \(P=30(10)+20(0)=300\)