zał. a,h > 0
Pole powierzchni calkowitej :
Pc = 2a2 + 4ah = 2a(a+h) = S (z polecenia)
| S | ||
zatem | − a = h | |
| 2a |
| S | 1 | |||
V(a) = a2 * ( | − a) = | aS − a3 | ||
| 2a | 2 |
| 1 | ||
V'(a) = | S − 3a2 | |
| 2 |
| 1 | |
S = 3a2 | |
| 2 |
| 1 | ||
a2 = | S | |
| 6 |
| √6 | ||
a = | √S | |
| 6 |
| √6 | ||
V(a) osiaga maksimum dla a = | √S | |
| 6 |
| 1 | √6 | √6 | ||||
i wynosi V(a) = | * | √S − ( | √S)3 = ... | |||
| 2 | 6 | 6 |