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2 years ago

If a and b are positive numbers, find the maximum value of f(x)=x^a*(1-x)^b, 0 less than or equal to x less than or equal to 1.

1 answer:

7 0

<span>f '(x) = ax^(a-1)(1 - x)^b - bx^a (1 - x)^(b-1) = x^(a-1)(1 - x)^(b-1) (a(1 - x) - bx).

If a < 1 or b < 1, this may not be defined at one or both of the end points. f '(x) = 0 if the right most term is zero.

a - ax - bx = 0 ==> x = a/(a + b).

Note that this number is both well define and is between 0 and 1.

f(0) = 0

f(1) = 0,

f(a/(a+b)) = a^ab^b/(a + b)^(a + b) <--- the maximum value.

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

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2 years ago

Let f(x,y,z) = ztan-1(y2) i + z3ln(x2 + 1) j + z k. find the flux of f across the part of the paraboloid x2 + y2 + z = 3 that li

Sophie [7]

Consider the closed region $V$ bounded simultaneously by the paraboloid and plane, jointly denoted $S$ . By the divergence theorem,

$\displaystyle\iint_S\mathbf f(x,y,z)\cdot\mathrm dS=\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV$

And since we have

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the volume integral will be much easier to compute. Converting to cylindrical coordinates, we have

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Then the integral over the paraboloid would be the difference of the integral over the total surface and the integral over the disk. Denoting the disk by $D$ , we have

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Parameterize $D$ by

$\mathbf s(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf j+2\,\mathbf k$
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which would give a unit normal vector of $\mathbf k$ . However, the divergence theorem requires that the closed surface $S$ be oriented with outward-pointing normal vectors, which means we should instead use $\mathbf s_v\times\mathbf s_u=-u\,\mathbf k$ .

Now,

$\displaystyle\iint_D\mathbf f\cdot\mathrm dS=\int_{u=0}^{u=1}\int_{v=0}^{v=2\pi}\mathbf f(x(u,v),y(u,v),z(u,v))\cdot(-u\,\mathbf k)\,\mathrm dv\,\mathrm du$
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So, the flux over the paraboloid alone is

$\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-(-2\pi)=\dfrac{5\pi}2$

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2 years ago

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