Question

We want to construct a box whose base length is 3 times the base width. The material used to build the top and bottom cost $10/ft^2 and the material used to build the sides cost $6/ft^2. If the box must have a volume of50ft^3 . determine the dimensions that will minimize the cost to build the box.

Answer 1

Answer:

Step-by-step explanation:

Let l be length, w be the width and h height of box.

given l = 3w.

Volume of box = lwh = 3w * w * h

=> 3w²h = 50  --------------- (1)

Let us now construct an expression for calculating cost of making the box.

C = 2(area of top/bottom)(10) + 2(area of front/back)(6) + 2(area of sides)(6)

    = 2(3w²)(10) + 2(3wh)(6) + 2(wh)(6)

    = 60w² + 36wh + 12wh

    = 60w² + 48wh ---------------- (2)

Now from (1) we can find out value of h in terms of w.

3w²h = 50 => h = 50/3w².

Substitute the value of h in (2)

C = 60w² + 48w (50/3w²)

   = 60w² + 800/w  ------------------ (3)

Now to calculate to minimum cost, differentiate w.r t to w.

C'  = 120w - 800/w² = (120w³ - 800) / w²  

Looking at the equation, minimum cost is possible when C' = 0 .

If C' = 0,

 0 = 120w³ - 800 / w²

=> 120w³ - 800 = 0

=> w³ = 800/120

=> w = ∛800/120 = 1.882 (approx.)

We can discard the C' at w = 0, because we cannot make a box when width is 0.

Now substituting the value of w in (1) to get the height of the box.

h = 50/3w² => 50 / 3 (1.882)² = 4.705 ft.

substituting the value of w in (3) to get the Cost of box

C = 60w² + 800/w = 60(1.882)² + 800/1.882

   =  $ 637.595

Dimensions of box are:

length = 5.646 ft., width = 1.882 ft, height = 4.705 ft.

Cost of box = $ 637.60

Answer 2

Let width of base =x
Length = 3 x
V= lb h = 50
X *3x h 50!
H= 50/3x^2
Cost
F(x)= 10*3x^2*2+6*2(x+3x) 50/3x^2
= 60x^2 + 800/x
F’(x)= 120x-800/x^2
(120x^3-800)/x^2
Putf’(x)=0
So x^3= 800/120=20/3
C=cube root 20/3
F”(x)= x=+ve
Minimum at x(20/3)^1/3