Answer:
Step-by-step explanation:
let side of cube=x cm
volume=x³ cm³
again side=(x-3) cm
volume=(x-3)³ cm³
x³-(x-3)³=1385
(a³-b³)=(a-b)(a²+ab+b²)
(x-x+3){x²+x(x-3)+(x-3)²}=1385
3(x^2+x²-3x+x²-6x+9)=1385
3(3x²-9x+9)=1385
9x²-27x+27=1385
9x²-27x+27-1385=0
9x²-27x-1358=0
Answer:
x = 2 cm
y = 2 cm
A(max) = 4 cm²
Step-by-step explanation: See Annex
The right isosceles triangle has two 45° angles and the right angle.
tan 45° = 1 = x / 4 - y or x = 4 - y y = 4 - x
A(r) = x* y
Area of the rectangle as a function of x
A(x) = x * ( 4 - x ) A(x) = 4*x - x²
Tacking derivatives on both sides of the equation:
A´(x) = 4 - 2*x A´(x) = 0 4 - 2*x = 0
2*x = 4
x = 2 cm
And y = 4 - 2 = 2 cm
The rectangle of maximum area result to be a square of side 2 cm
A(max) = 2*2 = 4 cm²
To find out if A(x) has a maximum in the point x = 2
We get the second derivative
A´´(x) = -2 A´´(x) < 0 then A(x) has a maximum at x = 2
