Maximize Q = Xy When X + Y = 70

Hey guys, today we're diving into a classic optimization problem that's super common in math, especially when you're first getting your feet wet with calculus or even just algebra. We're going to figure out the absolute maximum value of the product Q = xy when we know that the sum of x and y is fixed at 70, meaning x + y = 70. This kind of problem pops up everywhere, from figuring out the largest area you can fence with a fixed amount of material to optimizing profits in business scenarios. It's all about finding that sweet spot where your product is as big as possible, given a limitation on its components.

So, let's break it down. We have two variables, x and y, and our goal is to maximize their product, Q = xy. The catch, or the constraint as we call it in math-speak, is that x + y must equal 70. This constraint is key because it links x and y together, meaning if you change one, the other has to adjust. Without this constraint, you could make x and y infinitely large, and Q would also be infinitely large, which isn't very interesting. The constraint is what gives us a definite maximum to find. Think of it like this: you have $70 to spend on two items, and you want to get the biggest 'bang for your buck' by multiplying their prices. How do you split that $70 to maximize your 'value score'? That's essentially what we're solving here.

There are a few ways to tackle this, and we'll go through the most common and intuitive ones. One method involves using substitution, which is a fantastic algebraic technique. Since we know x + y = 70, we can easily express one variable in terms of the other. For instance, we can rewrite the constraint as y = 70 - x. Now, this is super handy because we can take this expression for y and plug it directly into our equation for Q. So, Q = xy becomes Q = x(70 - x). See what we did there? We've successfully transformed a problem with two variables (x and y) into a problem with just one variable (x). This is a huge step because it's much easier to find the maximum of a function of a single variable. After the substitution, our equation for Q looks like Q = 70x - x^2. This is a quadratic equation, and its graph is a parabola. Since the coefficient of the x^2 term is negative (-1), the parabola opens downwards, which means it has a distinct maximum point at its vertex. Finding the vertex of this parabola will give us the maximum value of Q.

To find the vertex of a parabola in the form ax^2 + bx + c, the x-coordinate of the vertex is given by the formula -b / (2a). In our case, Q = -x^2 + 70x, so a = -1 and b = 70. Plugging these values into the formula, we get x = -70 / (2 * -1) = -70 / -2 = 35. So, the maximum value of Q occurs when x is 35. Now, because we know x + y = 70, if x = 35, then y must also be 70 - 35 = 35. So, it turns out that x and y are equal when Q is maximized. This is a really common theme in optimization problems with symmetric constraints – the optimal solution often involves the variables being equal.

Once we have x = 35 and y = 35, we can easily calculate the maximum value of Q. Just plug these values back into the original Q = xy equation: Q = 35 * 35. This calculation gives us Q = 1225. So, the maximum value of Q is 1225, and it occurs when x = 35 and y = 35. Pretty neat, right? It really highlights how a simple constraint can lead to a very specific, optimal outcome. We've used basic algebra and the properties of quadratic functions to solve this, and it's a super powerful approach that you'll see used again and again in various math and science contexts. It's all about transforming your problem into a form you can analyze easily, and substitution is your best friend for that!

Using Calculus for Optimization

Alright, let's switch gears and talk about another super popular way to solve this exact same problem: using calculus! If you're into derivatives, this method is going to feel really natural. Remember how we got Q = 70x - x^2 after substituting y = 70 - x into Q = xy? Well, calculus gives us a direct way to find the maximum (or minimum) of a function. The core idea is that at the peak of a function (its maximum or minimum point), the slope of the tangent line is exactly zero. And what do we use to find the slope of a tangent line? That's right, the derivative!

So, we have our function Q(x) = 70x - x^2. To find where the maximum occurs, we need to find the derivative of Q with respect to x, denoted as dQ/dx or Q'(x). Taking the derivative of 70x is straightforward – it's just 70. And the derivative of -x^2 is -2x. So, our derivative is Q'(x) = 70 - 2x. Now, to find the critical points where a maximum or minimum might occur, we set this derivative equal to zero and solve for x:

70 - 2x = 0

Solving for x, we add 2x to both sides: 70 = 2x. Then, we divide both sides by 2: x = 70 / 2 = 35.

This value, x = 35, is our critical point. It's where the slope of the function Q(x) is zero. But is it a maximum or a minimum? In this case, we know from the quadratic form (-x^2) that it's a maximum, but in more complex scenarios, you'd use the second derivative test. The second derivative, Q''(x), is the derivative of Q'(x). Since Q'(x) = 70 - 2x, its derivative Q''(x) is simply -2. Because the second derivative is negative (-2 < 0), this confirms that our critical point at x = 35 corresponds to a local maximum. Since our function is a simple downward-opening parabola, this local maximum is also the absolute maximum.

Just like with the algebraic method, once we find x = 35, we can easily find y using our constraint x + y = 70. So, 35 + y = 70, which means y = 70 - 35 = 35. And again, we find that x and y are equal at the maximum. Finally, we calculate the maximum value of Q by plugging x = 35 and y = 35 back into Q = xy: Q = 35 * 35 = 1225.

So, using calculus, we arrive at the exact same answer: the maximum value of Q is 1225, occurring when x = 35 and y = 35. This demonstrates the power and versatility of calculus in solving optimization problems. It's a robust tool that works for a much wider range of functions than just simple quadratics.

Intuitive Understanding: The Power of Equality

Now, let's think about this problem in a more intuitive way, without getting too bogged down in formulas. We want to maximize Q = xy subject to x + y = 70. What does this really mean? We're looking for two numbers that add up to 70 and whose product is as large as possible. Let's try some pairs:

• If x = 1, then y = 69. Q = 1 * 69 = 69.
• If x = 10, then y = 60. Q = 10 * 60 = 600.
• If x = 20, then y = 50. Q = 20 * 50 = 1000.
• If x = 30, then y = 40. Q = 30 * 40 = 1200.
• If x = 35, then y = 35. Q = 35 * 35 = 1225.
• If x = 40, then y = 30. Q = 40 * 30 = 1200.
• If x = 50, then y = 20. Q = 50 * 20 = 1000.

As you can see from these examples, the product Q gets larger as x and y get closer to each other. When x and y are far apart (like 1 and 69), the product is small. As they move closer together, the product increases. The maximum value clearly occurs when x and y are as close as possible, which in this case, since their sum is an even number, means they are equal. So, x = y = 35 gives us the maximum product.

This principle – that for a fixed sum, the product is maximized when the numbers are equal – is incredibly powerful and shows up in many areas of mathematics and economics. It's related to the Arithmetic Mean-Geometric Mean (AM-GM) inequality, which states that for non-negative numbers, the geometric mean is always less than or equal to the arithmetic mean. In our case, the arithmetic mean of x and y is (x + y) / 2 = 70 / 2 = 35. The geometric mean is sqrt(xy). So, sqrt(xy) <= (x + y) / 2. Substituting the known values, we get sqrt(Q) <= 35. Squaring both sides gives us Q <= 35^2, which means Q <= 1225. The equality holds (meaning the maximum is achieved) precisely when x = y. This gives us a very elegant and quick way to solve such problems if you're familiar with the AM-GM inequality.

So, whether you use substitution and quadratic properties, calculus with derivatives, or the intuitive understanding backed by inequalities, the conclusion is the same. The maximum value of Q = xy when x + y = 70 is 1225, and this maximum is achieved when x = 35 and y = 35. It’s a beautiful illustration of how constraints shape outcomes and how simple mathematical principles can unlock complex-seeming problems. Keep practicing these, guys, because they build a really strong foundation for more advanced math!

Conclusion

To wrap things up, we've explored three fantastic methods to find the maximum value of Q = xy given the constraint x + y = 70. We saw how algebraic substitution transforms the problem into analyzing a downward-opening parabola, revealing the maximum at its vertex. Then, we used calculus, specifically derivatives, to find the point where the function's slope is zero, confirming a maximum. Finally, we touched upon the intuitive principle that for a fixed sum, the product is maximized when the numbers are equal, a concept related to the AM-GM inequality.

In every case, the result is consistent: the maximum value of Q is 1225, achieved when x = 35 and y = 35. This problem is a cornerstone for understanding optimization, a skill vital in countless fields. Understanding these approaches equips you with powerful tools for tackling similar challenges, whether in exams or real-world applications. Keep experimenting and exploring the elegance of mathematics!