Respuesta :
Answer:
a) V(x) = x ( 60 -2x )( 15-2x )
b) 0.55 inches ≤ x ≤ 6.79 inches
c) x ≥ 6.79 inches
Step-by-step explanation:
Given data:
No top, cardboard dimensions ; 15-in by 60-in
a) A function for the volume of the box as a function of x the Volume can be represented by this function below
= V(x) = x ( 60 -2x )( 15-2x )
where : x = height , ( 60 - 2x ) = length , ( 15 -2x ) = width
b) determine x so that the volume of the box ≥ 450 inches
450 = x( 60 - 2x ) ( 15 -2x ) ( solving the equation )
0.55 inches ≤ x ≤ 6.79 inches
c ) The value of x for which volume of the box is maximum
will be x ≥ 6.79 inches
The volume of the box as a function of x V(x) = x ( 60 -2x )( 15-2x )
The volume of the box as a function of x inches 0.55 inches ≤ x ≤ 6.79
The volume of the box is maximum x ≥ 6.79 inches
Given ,
The box with no top that is to be made by removing squares of width x
The corners of a 15-in by 60-in piece of cardboard.
- Volume can be represented by this function below
V(x) = x ( 60 -2x )( 15-2x )
Where : x = height , ( 60 - 2x ) = length , ( 15 -2x ) = width
The volume of the box as a function of x is V(x) = x ( 60 -2x )( 15-2x )
- To determine x so that,
The volume of the box ≥ 450 inches
V(x) = x ( 60 -2x )( 15-2x )
The volume of the box is at least 450 cubic inches.0.55 inches ≤ x ≤ 6.79 inches
- The value of x for which volume of the box is maximum will be x ≥ 6.79 inches.
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