24 uždavinys
1.
Sprendimas.
Ritinio tūris $$V = \pi\cdot r^{2}\cdot h$$, r = 6, aukštis h = 2x, nes vandens paviršius liečia rutuliuką.
Ritinio tūris $$V_{rit} = \pi\cdot 6^{2}\cdot 2\cdot x = 72\cdot \pi\cdot x$$.
Rutuliuko tūris yra $$\frac{4}{3}\cdot \pi\cdot x^{3}$$.
Vandens tūris $$72\cdot \pi\cdot x-\frac{4}{3}\cdot \pi\cdot x^{3}$$
2. Koks turi būti rutuliuko spindulio ilgis x, kad taip įpilto į indą vandens tūris būtų didžiausias?
Sprendimas.
Funkcijos $$72\cdot \pi\cdot x-\frac{4}{3}\cdot \pi\cdot x^{3}$$ išvestinę prilyginsime nuliui:
( 72* π
* x-
* π* x^3)′ =
0
* x- | 4 |
| / 3 |
* x^3)′ = 0
|
|
( 72* π* x-
* π* x^3)′ = 0$$(72\cdot \pi\cdot x-\frac{4}{3}\cdot \pi\cdot x^{3})'$$ = $$0$$
* x- | 4 |
| / 3 |
* x^3)′ = 0$$(72\cdot \pi\cdot x-\frac{4}{3}\cdot \pi\cdot x^{3})'$$ = $$0$$$${\normalsize (72\cdot \pi\cdot x-\frac{4}{3}\cdot \pi\cdot x^{3})'}$$ = $${\normalsize (72\cdot \pi\cdot x)'-(\frac{4}{3}\cdot \pi\cdot x^{3})'}$$
Paaiškinimas:
Paaiškinimas: Sumos išvestinė (f+g)′ = f′ + g′
( 72* π* x)′- (
* π* x^3)′ = 0$$(72\cdot \pi\cdot x)'-(\frac{4}{3}\cdot \pi\cdot x^{3})'$$ = $$0$$
* x)′- ( | 4 |
| / 3 |
* x^3)′ = 0$$(72\cdot \pi\cdot x)'-(\frac{4}{3}\cdot \pi\cdot x^{3})'$$ = $$0$$$${\normalsize (72\cdot \pi\cdot x)'}$$ = $${\normalsize 72\cdot \pi}$$ Paaiškinimas:
Paaiškinimas: x išvestinė yra 1
72* π- (
* π* x^3)′ = 0$$72\cdot \pi-(\frac{4}{3}\cdot \pi\cdot x^{3})'$$ = $$0$$
- ( | 4 |
| / 3 |
* x^3)′ = 0$$72\cdot \pi-(\frac{4}{3}\cdot \pi\cdot x^{3})'$$ = $$0$$$${\normalsize (\frac{4}{3}\cdot \pi\cdot x^{3})'}$$ = $${\normalsize 4\cdot \pi\cdot x^{2}}$$ Paaiškinimas:
Paaiškinimas: Laipsnio išvestinė, kur n = 3
72* π- 4* π* x^2 = 0$$72\cdot \pi-4\cdot \pi\cdot x^{2}$$ = $$0$$
- 4* π* x^2 = 0$$72\cdot \pi-4\cdot \pi\cdot x^{2}$$ = $$0$$ 72* π = 0+ 4* π* x^2$$72\cdot \pi$$ = $$0+4\cdot \pi\cdot x^{2}$$
= 0+ 4* π* x^2$$72\cdot \pi$$ = $$0+4\cdot \pi\cdot x^{2}$$ 72* π = 4* π* x^2$$72\cdot \pi$$ = $$4\cdot \pi\cdot x^{2}$$
= 4* π* x^2$$72\cdot \pi$$ = $$4\cdot \pi\cdot x^{2}$$| 72* π |
| / 4 |
* x^2$$\frac{72\cdot \pi}{4}$$ = $$\pi\cdot x^{2}$$| 72* π |
| / 4/ π |
$${\normalsize \frac{72\cdot \pi}{4\cdot \pi}}$$ = $${\normalsize 18}$$
18 = x^2$$18$$ = $$x^{2}$$
saknis(18) = saknis( x^2)$$\sqrt {18}$$ = $$\sqrt {x^{2}}$$
18) = saknis( x^2)$$\sqrt {18}$$ = $$\sqrt {x^{2}}$$$${\normalsize \sqrt {x^{2}}}$$ = $${\normalsize x}$$
saknis(18) = x$$\sqrt {18}$$ = $$x$$
18) = x$$\sqrt {18}$$ = $$x$$$${\normalsize \sqrt {18}}$$ = $${\normalsize 3\cdot \sqrt {2}}$$
3* saknis(2) = x$$3\cdot \sqrt {2}$$ = $$x$$
2) = x$$3\cdot \sqrt {2}$$ = $$x$$$$(72\cdot \pi\cdot x-\frac{4}{3}\cdot \pi\cdot x^{3})'$$ = $$0$$
$$(72\cdot \pi\cdot x)'-(\frac{4}{3}\cdot \pi\cdot x^{3})'$$ = $$0$$
$$72\cdot \pi-(\frac{4}{3}\cdot \pi\cdot x^{3})'$$ = $$0$$
$$72\cdot \pi-4\cdot \pi\cdot x^{2}$$ = $$0$$
$$72\cdot \pi$$ = $$4\cdot \pi\cdot x^{2}$$
$$\frac{72\cdot \pi}{4\cdot \pi}$$ = $$x^{2}$$
$$18$$ = $$x^{2}$$
$$\sqrt {18}$$ = $$\sqrt {x^{2}}$$
$$3\cdot \sqrt {2}$$ = $$x$$
Atsakymas:$$3\cdot \sqrt {2}$$