Respuesta :
Answer:
a. It should look for 3 linear factors.
b. There are 3 dimensions.
c. x = 1.7
d. Maximum value is 12.597
Step-by-step explanation:
a.
[tex]-x^{3} - x^{2} + 12 \timesx[/tex] = [tex]-x^{3} - 4 \timesx^{2} + 3 \timesx^{2} +12 \timesx[/tex] = [tex]-(x + 4) \timesx^{2} + 3 \timesx \times(x + 4)[/tex] = [tex](x + 4) \timesx \times (x + 3)[/tex]
Hence the factorized form of the volume of the poly is [tex](x + 4) \timesx \times (x + 3)[/tex]
3 linear factors should needed to be look for.
b.
The 3 dimensions are (x + 4), x, (x + 3).
c.
Let, f(x) = [tex]-x^{3} - x^{2} + 12 \timesx[/tex]
Now, [tex]\frac{d f(x)}{dx}[/tex] = [tex]-3 \timesx^{2} -2 \timesx + 12[/tex]
Let [tex]f1(x)[/tex] = [tex]-3 \timesx^{2} -2 \timesx + 12[/tex] and [tex]f2(x)[/tex] = [tex]\frac{d f1(x)}{dx}[/tex] = [tex]-3 \timesx -2[/tex]
The value will be maximum, if f1(x) = 0 and f2(x) < 0.
[tex]f1(x) = 0[/tex]\\ [tex]3 \timesx^{2} + 2 \timesx -12 = 0[/tex]\\ [tex]x = \frac{- 2 + \sqrt{2^{2} + 4 \times3 \times12} }{2 \times3} , \frac{- 2 - \sqrt{2^{2} + 4 \times3 \times12} }{2 \times3}[/tex]
x = 1.69 ≅ 1.7 or x = -2.36 ≅ -2.4
Now f2(1.7) < 0.
At x = 1.7, the volume will be maximum.
d.
The maximum value is f(1.7) = 12.597(approximately)
