Maximum Problem
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11-09-2021, 12:39 AM
Post: #1
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Maximum Problem
I’m wondering if solve app can be used to determine the value of x that is a maximum in producing a value of 1 for the integral from 0 to x of a^X. It appears as if a^~1.72 is the maximum. Integral from 0 to x of a^3 reaches 1 at ~1.414 (sqrt 2). Using a^1.72, the integral totals 1 when x reaches ~1.4447. Using a^1 the integral reaches 1 when x once again reaches ~1.414 (sqrt 2).
If not achievable in solve, how might I approach this problem? (Not for any purpose, just found myself curious when figuring out how to use the solve app with integrals!) |
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11-17-2021, 02:45 AM
Post: #2
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RE: Maximum Problem
(11-09-2021 12:39 AM)lrdheat Wrote: I’m wondering if solve app can be used to determine the value of x that is a maximum in producing a value of 1 for the integral from 0 to x of a^X. It appears as if a^~1.72 is the maximum. Integral from 0 to x of a^3 reaches 1 at ~1.414 (sqrt 2). Using a^1.72, the integral totals 1 when x reaches ~1.4447. Using a^1 the integral reaches 1 when x once again reaches ~1.414 (sqrt 2). I’ve read over your post a few times; I’m not 100% sure, but it seems to me the constraint you’re wanting satisfied is \[ \int_0^X a^Y \mathrm{d}a = 1, \] which is equivalent to (via integration) \[ \frac{X^{(Y+1)}}{Y+1} = 1, \] over the domain I’m assuming is of interest (non-negative X and Y). (This agrees with your observations regarding the quadrant I curve for Y=1 and Y=3.) I’m attaching a plot of this, with the definition showing, as well as a screenshot of the coordinates found when selecting PoI / Horizontal Extrema as the Trace mode (with the cursor near the horizontal extrema in quadrant I), both from the Advanced Graphing app. (The reported coordinates are quite close to the corresponding numbers from your post: 1.71828182846 ≈ 1.72 and 1.44466786101 ≈ 1.4447, so it seems likely I’ve gotten the gist of your post’s stated mathematical problem.) |
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11-17-2021, 06:17 AM
Post: #3
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RE: Maximum Problem
Cool!
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