Question

A jewel box is to be constructed of materials that costs K1 per square inch for the bottom, K2 per square inch for the sides, and K5 per square inch for the top. If the total volume is to be 96 inch cubic, advise Mr Lukonde on what dimensions will minimize the total cost of construction. Show all the calculations.

Answer 1

Answer:

x= 5,77 *∛ K₂ / ( K₁ + K₅ ) the side f the square bottom-top

h = 2,88/ [∛ K₂ / ( K₁ + K₅ ) ]² heigh of the box

Step-by-step explanation:

Data from problem statement only gives one relation between dimensions of the box, we need two, in order to express surface area as a function of just one variable. In such a case we must assume the box is of square bottom and top.

Then

Area of the bottom                      A(b) = x²          ⇒  C(b) = K₁*x²

Area of the top                              A(t)  = x²         ⇒  C(t)  = K₅*x²

Total lateral area ( 4 sides)          A(l)  = 4*x*h   ⇒   C(l)  = 4*K₂*x*h

V(bx) = 96 in³

V(bx) = x²*h = 96

h = 96/x²

Then total cost as a function of x

C(x) =  K₁*x² +  K₅*x² + 4*K₂*(96)/x

Taking derivatives on both sides of the equation

C´(x) = 2*K₁*x + 2*K₅*x - 384*K₂/x²

C´(x) = 0         ⇒     2*K₁*x + 2*K₅*x - 384*K₂/x² = 0

2*K₁*x³ + 2*K₅*x³ = 384*K₂       or      K₁*x³ + k₅*x³ =  192*K₂

x³ ( K₁ + K₅ ) = 192*K₂

x  = ∛192*K₂ / ( K₁ + K₅ )

x= 5,77 *∛ K₂ / ( K₁ + K₅ )

If we obtain the second derivative

C´´(x)  = 2*K₁ + 2*K₅ - (-2*x)*384*K₂/x⁴

C´´(x)  = 2*K₁ + 2*K₅ + 768*K₂/x³

As x can not be negative  the expression C´´ wil be  C´´> 0

then we have a minimum for the function for x = 5,77 *∛ K₂ / ( K₁ + K₅ )

and  h = 96 / [ 5,77 *∛ K₂ / ( K₁ + K₅ )]²

h = 2,88/ [∛ K₂ / ( K₁ + K₅ ) ]²