Among all pairs of numbers whose sum is 100, how do you find a pair whose product is as large as possible. (Hint: express the product as a function of x)?

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Among all pairs of numbers whose sum is 100, how do you find a pair whose product is as large as possible. (Hint: express the product as a function of x)?

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Answer 1 · 2016-12-10T14:15:41

$50, 50$ $50, 50$

Explanation:

Suppose two numbers sum to equal $100$ $100$ . Let $x$ $x$ represent the first number. Then the second number must be $100 - x$ $100-x$ , and their product must be $x (100 - x) = - x^{2} + 100 x$ $x(100-x) = -x^2+100x$ .

As $f (x) = - x^{2} + 100 x$ $f(x)=-x^2+100x$ is a downward opening parabola, it has a maximum at its vertex. To find its vertex, we put it in vertex form, that is, $a {(x - h)}^{2} + k$ $a(x-h)^2+k$ where $(x, f (x)) = (h, k)$ $(x,f(x))=(h, k)$ is its vertex.

To put it into vertex form, we use a process called completing the square:

$- x^{2} + 100 x = - (x^{2} - 100 x)$ $-x^2+100x = -(x^2-100x)$

$= - (x^{2} - 100 x) - {(\frac{100}{2})}^{2} + {(\frac{100}{2})}^{2}$ $=-(x^2-100x)-(100/2)^2+(100/2)^2$

$= - (x^{2} - 100 x) - 2500 + 2500$ $=-(x^2-100x)-2500+2500$

$= - (x^{2} - 100 x + 2500) + 2500$ $=-(x^2-100x+2500)+2500$

$= - {(x - 50)}^{2} + 2500$ $=-(x-50)^2+2500$

Thus the vertex is at $(x, f (x)) = (50, 2500)$ $(x, f(x)) = (50, 2500)$ , meaning it attains a maximum of $2500$ $2500$ when $x = 50$ $x=50$ .

As such, the pair of numbers $x, 100 - x$ $x, 100-x$ attains a maximal product when $x = 50$ $x=50$ , meaning the desired pair is $50, 100 - 50$ $50, 100-50$ , or $50, 50$ $50, 50$ .

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Among all pairs of numbers whose sum is 100, how do you find a pair whose product is as large as possible. (Hint: express the product as a function of x)?

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