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To find the numbers that maximize the product p, we can use the formula for a quadratic equation: x = -b / 2a. Let's call one number x and the other number (120-x). Therefore, the equation becomes x(120 - x^2), which simplifies to -x^3 + 120x. We can find x by setting the derivative equal to zero, resulting in x = 10. Therefore, the two numbers that maximize the product are 10 and 110, with a product of 12100.

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5mo ago
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14y ago

802 x 40 = 256000

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Q: What are two numbers whose sum is 120 such that the product p of one number and the square of the remaining number is a maximum?
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