The Purplemath Forums

by **nona.m.nona** on Fri Jan 30, 2009 2:30 am

For what values of the numbers $a$ and $b$ does the function $f(x)\, =\, axe^{bx^2}$ have the maximum value $f(2)\, =\, 1 \mbox{?}$

$f(2)\, =\, a(2)e^{b(2)^2}\, =\, 2ae^{4b}\, =\, 1$

We've done the derivative tests.

$f'(x)\, =\, ae^{bx^2}\, +\, ax\left(2bx\right) e^{bx^2}\, =\, 0$

$ae^{bx^2}\left(1\, +\, 2bx^2\right)\, =\, 0$

Plug 2 in for x.

$ae^{bx^2}\, =\, 0\, \mbox{ or }\, 1\, +\, 8b\, =\, 0$

$a\, =\, 0\, \mbox{ or } \, b\, =\, -\frac{1}{8}$

But if $a\, =\, 0$ , then $f(x)$ would be always zero.

$f''(x)\, =\,a\left(2bx\right)e^{bx^2}\, +\, 4abxe^{bx^2}\, +\, 2abx^2 \left(2bx\right) e^{bx^2}$

$=\, 2abxe^{bx^2}\, +\, 4abxe^{bx^2}\, +\, 4ab^2 x^3 e^{bx^2}$

$=\, 2abxe^{bx^2}\left(3\, +\, 2bx^2\right)\, =\, 0$

Plug 2 in for x.

$4abe^{4b}\left(3\, +\, 8b\right)\, <\, 0$

For the product to be less than zero (so the function is concave down giving a max), the factors have to have opposite signs.

$4abe^{4b}\, >\, 0\, \mbox{ and } \, 3\, +\, 8b\, <\, 0$

$ab\, >\, 0\, \mbox{ and } \, b\, <\, -\frac{3}{8}$

But $b\, =\, -\frac{1}{8}\, >\, -\frac{3}{8}$ so $ab\, <\, 0\, \mbox{ and }\, b\, >\, -\frac{3}{8}$

Back at the start:

$f(2)\, =\, a(2)e^{b(2)^2}\, =\, 2ae^{4b}\, =\, 1$

$2ae^{-\frac{1}{2}}\, =\, 1$

$\frac{2a}{\sqrt{e}}\, =\, 1$

$a\, =\, \frac{\sqrt{e}}{2}$

Is that right? Thx.

**Sponsor**

by **stapel_eliz** on Fri Jan 30, 2009 6:24 pm

You can always check your work:

. . . . . $f(x)\, =\, \left(\frac{\sqrt{e}}{2}\right) x e^{-\frac{x^2}{8}}$

Does (2, 1) lie on this curve?

. . . . . $f(2)\, =\, \left(\frac{\sqrt{e}}{2}\right)\left(2\right) e^{-\frac{4}{8}}\, =\,\left(\frac{\sqrt{e}}{2}\right) \left(\frac{2}{1}\right) \left(\frac{1}{\sqrt{e}}\right)\, =\, 1$

Is there a critical point at x = 2?

. . . . . $f'(x)\, =\, \left(\frac{\sqrt{e}}{2}\right) e^{-\frac{x^2}{8}}\, +\, \left(\frac{\sqrt{e}}{2}\right) x \left(-\frac{x}{4}\right) e^{-\frac{x^2}{8}}$

. . . . . $f'(2)\, =\, \left(\frac{\sqrt{e}}{2}\right) e^{-\frac{1}{2}}\, +\, \left(\frac{\sqrt{e}}{2}\right) \left(\frac{2}{1}\right) \left(-\frac{1}{2}\right) e^{-\frac{1}{2}}$

. . . . . . . . .. $\, =\, \left(\frac{\sqrt{e}}{2}\right) e^{-\frac{1}{2}}\, -\, \left(\frac{\sqrt{e}}{2}\right) e^{-\frac{1}{2}}\, =\, 0$

And is the second derivative negative at x = 2, so the curve is concave down, and the critical point represents a maximum?

. . . . . $f''(x)\, =\, \left(-\frac{\sqrt{e}x}{8}\right) \left(e^{-\frac{x^2}{8}}\right) \left(3\, -\, \frac{x^2}{4}\right)$

. . . . . $f''(2)\, =\, \left(-\frac{\sqrt{e}}{4}\right) \left(\frac{1}{\sqrt{e}}\right) \left(3\, -\, 1\right)$

. . . . . . . . . .. $=\,\left(-\frac{1}{4}\right) \left(2\right)\, =\, -\frac{1}{2}\, \leq\, 0$

It looks good to me! :thumb:

Eliz.

The Purplemath Forums

values where f(x) = axe^(bx^2) has max f(2) = 1

values where f(x) = axe^(bx^2) has max f(2) = 1