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 :)
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
To solve this problem, we'll use the formulas for the volume and surface area of a rectangular box.
Given that the sides of the box have lengths 2x, 2x, and y, the volume of the box can be expressed as:
V = length * width * height
64 = 2x * 2x * y
Simplifying this equation, we get:
64 = 4x^2 * y
Now let's solve the parts of the problem one by one:
A) To find the surface area (A) of the box, we need to find the area of each face and then sum them up. Since the box has six faces, we have:
A = 2xy + 2(2x)(y) + 2(2x)(2x)
A = 2xy + 4xy + 8x^2
A = 6xy + 8x^2
B) To find the value of x for which dA/dx = 0, we need to differentiate the equation for A with respect to x and set it equal to zero. Let's differentiate the equation:
dA/dx = 6y + 16x
Setting this equal to zero:
6y + 16x = 0
16x = -6y
x = -6y/16
x = -3y/8
C) To find d^2A/dx^2, we need to differentiate the equation for dA/dx with respect to x. Differentiating dA/dx, we get:
d^2A/dx^2 = d/dx (dA/dx)
d^2A/dx^2 = d/dx (6y + 16x)
d^2A/dx^2 = 16
Since the second derivative is constant (16), it means that x does not affect whether the value obtained in part B gives a maximum or minimum. We need to determine this based on the sign of the first derivative.
If the first derivative, dA/dx, is positive, then the value of x obtained in part B gives a minimum. If dA/dx is negative, then the value of x obtained in part B gives a maximum.
D) To find the maximum or minimum value of A, we need to substitute the value of x obtained in part B back into the equation for A. Substituting x = -3y/8 into A, we get:
A = 6xy + 8x^2
A = 6(-3y/8)y + 8(-3y/8)^2
A = -9y^2/4 + 9y^2/4
A = 0
Therefore, the maximum or minimum value of A is 0.
Hope this helps! Let me know if you have any further questions.