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Algebra Play - The Largest Product

Grade 8CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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When the sum of two numbers is constant, say x+y=Sx + y = S, their product xyxy is maximum when the two numbers are equal, i.e., x=y=S2x = y = \frac{S}{2}.

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If the sum SS is an odd integer and we are restricted to using integers, the maximum product is achieved when the numbers are as close as possible, differing by only 11.

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Algebraically, the relationship between the sum and the product can be seen in the identity: xy=(x+y2)2−(x−y2)2xy = \left( \frac{x+y}{2} \right)^2 - \left( \frac{x-y}{2} \right)^2. Since the first term (x+y2)2\left( \frac{x+y}{2} \right)^2 is constant for a fixed sum, the product xyxy is largest when the second term (x−y2)2\left( \frac{x-y}{2} \right)^2 is smallest, which happens when x−y=0x - y = 0 or x=yx = y.

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For a fixed perimeter of a rectangle, the area (which is the product of length and breadth) is maximum when the rectangle is a square (length=breadthlength = breadth).

📐Formulae

(x+y)2=(x−y)2+4xy(x + y)^2 = (x - y)^2 + 4xy

xy=(x+y2)2−(x−y2)2xy = \left( \frac{x+y}{2} \right)^2 - \left( \frac{x-y}{2} \right)^2

Maximum Product (xy) occurs when x=y\text{Maximum Product } (xy) \text{ occurs when } x = y

For a fixed sum S, Max Product =(S2)2\text{For a fixed sum } S, \text{ Max Product } = \left( \frac{S}{2} \right)^2

💡Examples

Problem 1:

Divide the number 1212 into two parts such that their product is the largest possible.

Solution:

Let the two parts be xx and yy. We are given x+y=12x + y = 12. To maximize xyxy, we set x=yx = y. Therefore, x=y=122=6x = y = \frac{12}{2} = 6. The maximum product is 6×6=366 \times 6 = 36.

Explanation:

If we chose other parts like 55 and 77, the product would be 3535. If we chose 44 and 88, the product would be 3232. Thus, 6×6=366 \times 6 = 36 is the largest.

Problem 2:

Divide 1515 into two integers such that their product is maximum.

Solution:

The sum is x+y=15x + y = 15. For maximum product, xx and yy should be as close as possible. Since 1515 is odd, we take x=7x = 7 and y=8y = 8. The maximum product is 7×8=567 \times 8 = 56.

Explanation:

For any two numbers with a fixed sum, the product decreases as the difference between the numbers increases. The closest possible integers for sum 1515 are 77 and 88.

Problem 3:

Prove using the identity xy=(x+y2)2−(x−y2)2xy = \left( \frac{x+y}{2} \right)^2 - \left( \frac{x-y}{2} \right)^2 that for x+y=20x + y = 20, the maximum value of xyxy is 100100.

Solution:

Substitute x+y=20x + y = 20 into the identity: xy=(202)2−(x−y2)2xy = \left( \frac{20}{2} \right)^2 - \left( \frac{x-y}{2} \right)^2 xy=102−(x−y2)2xy = 10^2 - \left( \frac{x-y}{2} \right)^2 xy=100−(x−y2)2xy = 100 - \left( \frac{x-y}{2} \right)^2 Since a squared term is always ≥0\ge 0, the maximum value of xyxy occurs when (x−y2)2=0\left( \frac{x-y}{2} \right)^2 = 0, which gives xy=100xy = 100.

Explanation:

This demonstrates that any difference between xx and yy will subtract a positive value from 100100, making the product smaller.