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7 respuestas

lo voy a hacer aki::

x+y = 12 donde y=12-x
donde f(x) = x*y^2 MAX

entonces f(x) = x(12-x) ^2

derivamos:: f'(x) = (12-x)^2 + 2 * x *(12-x)*(-1)

= 144-24x+x^2 -24x +2x^2
= 3x^2 -48x +144=0
= 3 ( x^2 - 16x + 48) =0
implica que
x = 12 ó x= 4

12 es mínimo
y 4 es máximo

entonces y= 8

su suma es 12 y su producto es 256






byebye

2007-01-27 06:18:16 · answer #1 · answered by Piet 3 · 1 1

y = 12 - x

x2 (12-x) = p(x)

p´(x) = 2x ( 12-x )+ x^2 (-1)= 24 x -2 x^2 -x^2 = -3 x^2 +24 x

p'(x) =0 -------> -3 x (x - 8) -----------> x = 0 ó x = 8

p''(x) = -6 x + 24

p''(0) = 24> 0 mínimo
p" (8) = - 48+24 <0 máximo

Los números son 8 y 4

2007-01-27 06:18:28 · answer #2 · answered by silvia g 6 · 2 1

4 y 8.

2007-01-27 06:05:21 · answer #3 · answered by Belmonte 4 · 1 0

2+10=12
10*10= 100

2007-01-30 15:29:35 · answer #4 · answered by chilipepper 4 · 0 1

-10 y +2

2007-01-27 10:29:54 · answer #5 · answered by N. R 5 · 0 1

9 + 3 =12
9 por 9 = 81

2007-01-27 06:12:25 · answer #6 · answered by H@d@ 20 2 · 0 2

6?

2007-01-27 06:07:17 · answer #7 · answered by rakelish 2 · 0 2

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