Minimium
Find the minimum value of the expression a − 5 + a − 4 + 3 a − 3 + a 8 + a 1 0 + 1 where a ∈ R +
The answer is 8.0.
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2 solutions
Quite Simple One.
The top black two lines have not been written by me there !!!! M y s o l u t i o n i s i n c o l o r s . S i n c e a l l t e r m s a r e ≥ 0 , w h e n w e a d d t h e m , a − 5 + a − 4 + 3 a − 3 + a 8 + a 1 0 + 1 > 0 . D i f f e r e n t i a t i n g a n d e q u a t i n g t o z e r o , f ′ ( a ) = a − 6 ∗ ( − 5 − 4 a − 9 a 2 + 8 a 1 3 + 1 0 a 1 5 ) = 0 . S i n c e a − 6 i s n o t e q u a l t o z e r o , − 5 − 4 a − 9 a 2 + 8 a 1 3 + 1 0 a 1 5 = 0 . ∵ c o e f f i c i e n t a d d u p t o 0 , a = 1 i s o n e s o l u t i o n . f " ( a ) = − 6 a − 7 ∗ ( − 5 − 4 a − 9 a 2 + 8 a 1 3 + 1 0 a 1 5 ) + a − 6 ∗ ( − 4 − 1 8 a + 1 0 4 a 1 2 + 1 5 0 a 1 4 ) . f " ( 1 ) = 0 + 2 3 2 > 0 , s o a = 1 i s t h e m i n i m u m s o l u t i o n . ∴ f ( 1 ) = 1 + 1 + 3 + 1 + 1 + 1 = 8 .
Modified after comments by Mr. Calvin Lin and Mr. James Wilson. Thanks to both.
How is it "very clear that this is only possible if a = 1 "?
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Since the sum of coefficients of Exp.=0, addes up to 0.
I have inserted this now. I hope it is correct now.
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That shows that is is possible. How is it "only possible"? IE Why aren't there any other real solutions to this degree 9 polynomial?
As James mentioned, we can use descartes' rule of signs to conclude that there is exactly one positive root. We also have to check the second derivative to ensure that it's a minimum (though that's reasonably obvious).
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@Calvin Lin – Thank you for detailed explanation. I have modified my solution accordingly.
Hi Niranjan. I'm just letting you know you made a mistake with your exponents. However, your idea behind the solution is still correct. Your method can be justified by applying Descartes's Rule of Signs. Since there is only one sign change in the polynomial, this tells us the polynomial has exactly one positive real root; and you found it (despite the exponent errors). Then justifying it's a minimum is easy because the second derivative has all positive terms.
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Thanks a lot. I have now made corrections . I wonder how I got such wrong exponents !
I fail to understand why there are top two linens there since I have not written them.
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I don't know. I still haven't figured out how to use LaTeX on this site yet.
Since all the terms are positive real, we can apply AM-GM inequality :
a − 5 + a − 4 + a − 3 + a − 3 + a − 3 + a 8 + a 1 0 + 1 ≥ 7 7 a 0 + 1 = 8 .
Equality occurs when a = 1 .