Arectangular box with a volume of 1088 ftcubed is to be constructed with a square base and top. the cost per square foot for the bottom is 15cents, for the top is 10cents, and for the sides is 2.5cents. what dimensions will minimize the cost?
what are the dimensions of the box? a. the length of one side of the base is ft? b. the hieght of the box is ?

x  = 11,09 ft

h  =  8,85  ft

Step-by-step explanation:

Box with square base and top

Let  x b a side of the square

A₁  (area of base)     A₁ =  x²

area of the top is equal to area of base   A₂  = x²

Cost of area A₁    

C₁  =  15 * x²

Cost of area  A₂

C₂  = 10 * x²

Area of the side of height  h

A₃ =  2*π*x*h                           V = 1088  = x²*h        h  = 1088/x²

A₃  = 2*π*x* (1088)/x²     A₃  =   6832,64/x

Cost  of A₃

C₃  = 2,5*4 *6832,64/x           C₃  = 68326,4/x

Total cost C

C(x)  = 15x²  +  10x²  +  68326,4/x

Taking derivatives on both sides of the equation

C´(x)   =  30 x  + 20 x  -   68326,4/x²

C´(x)   =  50 x  - 68326,4 /x²        C´(x)   = 0

50 x  - 68326,4 /x²   = 0

50x³    =   68326,4

x³   =   1366,53

x  = 11,09 ft     and the  height     h  =  1088/x²

h  =  8,85  ft

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step-by-step explanation:

x coordinate can be written by using the formula, x= x1 + 2/5(x2 - x1)

step-by-step explanation:

formula

x = x1 + p/w(x2 - x1)

let the line segment ab with a(x1,y1) and b(x2,y2)

a point(x,y) divide ab in the ratio, 2: 3 the x coordinate can be written as

x = x1 + p/w(x2 - x1)

here p=2 and w = 2+3

therefore,

x coordinate can be written by using the formula, x= x1 + 2/5(x2 - x1)

similarly y coordinate y = y1 + 2/5(y2 - y1)