Solve Systems Graphically and Interpret Intersections - College Algebra

1. A farmer has 100 feet of fencing to create a rectangular pen. If the length is $y$ feet and the width is $x$ feet, the constraint equation is $2x + 2y = 100$. If the farmer also wants the area $A = xy$ to equal 600 square feet, at what dimensions do these constraints intersect?
   Answer: $(20, 30)$ and $(30, 20)$ - two possible rectangular configurations When you encounter problems involving multiple constraints, you're looking for points where two equations intersect - this is a classic system of equations problem. You have two constraints: the perimeter constraint $2x + 2y = 100$ (which simplifies to $x + y = 50$) and the area constraint $xy = 600$. To find where these intersect, substitute the first equation into the second. From $x + y = 50$, you get $y = 50 - x$. Substituting into $xy = 600$: $x(50 - x) = 600$ $50x - x^2 = 600$ $x^2 - 50x + 600 = 0$ Using the quadratic formula: $x = \frac{50 \pm \sqrt{2500 - 2400}}{2} = \frac{50 \pm 10}{2}$ This gives $x = 30$ or $x = 20$. When $x = 20$, $y = 30$; when $x = 30$, $y = 20$. So the intersections are $(20, 30)$ and $(30, 20)$. Answer A is wrong because $(25, 25)$ gives an area of only 625 square feet, not 600. Answer B is incorrect because $(10, 40)$ and $(40, 10)$ satisfy the perimeter constraint but give areas of 400 square feet, not 600. Answer D is wrong because we just proved intersections do exist. Study tip: When solving constraint problems, always check that your solutions satisfy both original equations. Systems with a linear and quadratic equation typically yield zero, one, or two intersection points.
2. Two investment accounts grow linearly over time. Account A has value $V_A = 1000 + 50t$ and Account B has value $V_B = 800 + 75t$, where $t$ is time in months and values are in dollars. When do the accounts have equal value, and what happens after this point?
   Answer: $t = 8$ months, $1400; Account B grows faster afterward Setting $V_A = V_B$: $1000 + 50t = 800 + 75t$. Solving: $200 = 25t$, so $t = 8$ months. At $t = 8$: $V_A = V_B = 1400$. After this point, Account B grows at $75 per month vs Account A's $50 per month, so B grows faster. Choice A has wrong growth comparison. Choice C has wrong intersection time. Choice D has wrong time and incorrect behavior.
3. A system of equations has lines with slopes $m_1 = 3$ and $m_2 = 3$, but different y-intercepts. When graphed, these lines will:
   Answer: Be parallel with no intersection points, making the system inconsistent Lines with identical slopes ($m_1 = m_2 = 3$) but different y-intercepts are parallel and never intersect. This creates an inconsistent system with no solution. Choice A would be correct if the slopes were different. Choice C would be correct if the lines were identical (same slope AND same y-intercept). Choice D is impossible since two lines can intersect at most once.
4. A company's revenue $R$ (in thousands of dollars) is given by $R = 8x$, where $x$ is the number of units sold (in hundreds). The company's costs $C$ (in thousands of dollars) follow $C = 3x + 15$. At what production level do the revenue and cost functions intersect, and what is the economic interpretation?
   Answer: $x = 3$ hundred units; break-even point where revenue equals cost at $24,000 To find the intersection, set $R = C$: $8x = 3x + 15$. Solving: $5x = 15$, so $x = 3$. At $x = 3$: $R = 8(3) = 24$ and $C = 3(3) + 15 = 24$. Both equal 24 thousand dollars, making this the break-even point. Choice B has wrong calculation. Choice C misinterprets intersection as optimization. Choice D confuses intersection with marginal analysis.
5. Two cell phone plans are compared: Plan A charges $25 per month plus $0.10 per minute, while Plan B charges $15 per month plus $0.15 per minute. If the total monthly costs are equal, how many minutes were used, and what is the significance of this point graphically?
   Answer: 200 minutes; this is where the two linear cost functions intersect, representing equal total cost When you encounter problems comparing two linear cost functions, you're looking for the intersection point where both plans cost the same amount. This requires setting up equations for each plan and solving for when they're equal. Let's set up the cost functions. Plan A: $C_A = 25 + 0.10m$, and Plan B: $C_B = 15 + 0.15m$, where $m$ is minutes used. To find when costs are equal, set the equations equal: $25 + 0.10m = 15 + 0.15m$. Subtracting $0.10m$ from both sides gives $25 = 15 + 0.05m$, then $10 = 0.05m$, so $m = 200$ minutes. Graphically, this represents the intersection point of two lines, where both plans cost exactly the same amount. Answer A incorrectly suggests this point relates to equal slopes, but the slopes ($0.10$ and $0.15$) are the per-minute rates and never change. Answer B gives the wrong number of minutes and misinterprets the significance—after 200 minutes, Plan B becomes more expensive than Plan A, not at 150 minutes. Answer C correctly identifies 200 minutes but incorrectly describes this as Plan B's minimum cost, when actually Plan B's minimum occurs at zero minutes (just the base fee). Answer D correctly identifies both the 200-minute solution and its graphical meaning: the intersection point of two linear functions where total costs are equal. Study tip: In cost comparison problems, always set up equations for each option and solve for equality. The intersection point tells you exactly when switching between options makes financial sense.

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