4 x^2 y = 64 so y = 16/x^2
I guess it has a square bottom and height y
Area = area top and bottom + 4 x y
= 2*4 x^2 + 4* 2 x y
= 8 x^2 + 8 x y
= 8 x^2 + 8 x (16/x^2)
= 8 x^2 + 128/x agreed
dA/dx = 16 x -128/x^2
= 0 at max or min
16 x^3 = 128
x^3 = 8
x = 2
c) d^2A/dx^2 = 16 +128(2x)/x^4
at x = 2
16 + positive = positive so headed up , MINIMUM
d) at x = 2
A = 8 x^2 + 128/x
= 32 + 64
A closed rectangular box is made with sides of length (in cm) 2x,2x and y respectively.
The volume of the box is 64cm^3.
A) show that the surface area (six faces) of the box is given by:
A = 8x^2+128/x
B)find the Valur of x for which dA/dx=0
C)find d^2A/dx^2 and use it to determine whether the value obtained in (b) gives a maximum or minimum.
D) find the max or min value of A
Please help
Thankyou :)
1 answer
1 answer