Question

An open-top rectangular box is being constructed to hold a volume of 150 in3. The base of the box is made from a material costing 5 cents/in2. The front of the box must be decorated and will cost 10 cents/in2. The remainder of the sides will cost 2 cents/in2. Find the dimensions that will minimize the cost of constructing this box.

Answer 1

Answer:

minimum cost of construction box are x = x = 3.4199 in and y = 10.2598 in and z = 4.2750 in

Step-by-step explanation:

given data

volume = 150 in³

base material costing = 5 cents/in²

front cost = 10 cents/in²

remainder sides cost = 2 cents/in²

to find out

the dimensions that will minimize the cost

solution

we consider here length = x and breadth = y and height = z

and

area of base = xy

area of front = xz

and area of remaining side = xz + 2yz     .....................1

so

cost of base will be = 5xy

cost of front = 10xz

cost of remaining side = 2 ( xz+ 2yz)        

and

total cost will be

total cost TC = 5xy + 10xz + 2 ( xz+ 2yz)  

total cost TC = 5xy + 10xz + 2xz+ 4yz

total cost TC = 5xy + 12xz + 4yz                  ....................2

and total volume will be = xyz

150 = xyz

z = $\frac{150}{xy}$                  .......................3

now put z value in equation 2

total cost TC = 5xy + 12xz + 4yz    

total cost TC = 5xy + 12x$\frac{150}{xy}$ + 4y$\frac{150}{xy}$

total cost TC = 5xy + $\frac{1800}{y}$ + $\frac{600}{x}$     ...........4

now differentiate TC w.r.t x and y

TC (x) = 5y - $\frac{600}{x^2}$

TC (x) = 5x - $\frac{1800}{y^2}$

now equating with 0 these

5y - $\frac{600}{x^2}$ = 0

x² = $\frac{120}{y}$    

and

5x - $\frac{1800}{y^2}$ = 0

x = $\frac{360}{y^2}$    

solve we get

y = 10.2598

x = 3.4199

now put x and y value in equation 3

z = $\frac{150}{xy}$  

z = $\frac{150}{3.4199*10.2598}$  

z = 4.2750

so minimum cost of construction box are x = x = 3.4199 in and y = 10.2598 in and z = 4.2750 in