Prove by contradiction that the interval (a, b.has no minimum element.

Mathematics

1 answer:

Snowcat [4.5K]2 years ago

8 0

Assume (a,b) has a minimum element m.

m is in the interval so a < m < b.

a < m

Adding a to both sides,

2a < a + m

Adding m to both sides of the first inequality,

a + m < 2m

2a < a+m < 2m

a < (a+m)/2 < m < b

Since the average (a+m)/2 is in the range (a,b) and less than m, that contradicts our assumption that m is the minimum. So we conclude there is no minimum since given any purported minimum we can always compute something smaller in the range.

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Anna11 [10]

Check the picture.

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so $f(x)=4 x^{3}-260x^{2}+4000x$

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solving the quadratic equation, either by a calculator, graphing software, or by other algebraic methods as the discriminant formula, we find the solutions

x=10 and x=33.333

plug in f(x) these values to see which greater:

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Answer: 18000 cm cubed