If an open box has a square base and a volume of 115 in^3. and is constructed fr

Question

If an open box has a square base and a volume of 115 in^3. and is constructed from a tin sheet, find the dimensions of the box, assuming a minimum amount of material is used in its construction. (Round your answers to two decimal places.) height in length in width in A rectangular box is to have a square base and a volume of 20 ft^3. If the material for the base costs $0.29 per square foot, the material for the sides costs $0.10 per square foot, and the material for the top costs $0.21 per square foot, determine the dimensions of the box that can be constructed at minimum cost. length ft width ft height ft

Explanation / Answer

(1)Let the square base have dimension of "x" and height of the box be "h". It is given that
V = 115 = x²h
=> h = 115/x²
Now Surface Area of the tin material will be,
S = x² + 4hx
S = x² + 460/x
For minimum surface area, dS/dx = 0
=> 2x - 460/x² = 0
=> x = 6.12 in.
test for minimum area by second derivative test which shouild be positive at the above value of "x2
d²S/dx² = 2 + 920/x³ > 0 @ (x = 6.12 in.) => Minima
Hence
x = 6.12 in.
h = 115/( 6.12 )²=
h = 3.07 in

(2)let, x = side of square base in feet
h = height of box in feet
cost of base = 0.29x^2
cost of 4 sides =0.4hx
cost of top = 0.21x^2
Therfore tota Cost C=0.29x^2+0.4xh+0.21x^2
volume of box = 20 ft^3=x^2h

=> h=20/x^2

Now C=0.29x^2+0.4xh+0.21x^2=0.5x^2+0.4x(20/x^2)=0.5x^2+8/x
For Minimum cost
C'(x)=0

=> 1x-8/x^2=0 => x^3=8 => x=2
C"(x)=16/x^3

=>C"(2)=2>0 hence minimum at x=2
therfore C(2)=0.5(2)^2+8/2=2+4=6

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