You want to make a rectangular box that is x cm high, (x+5) cm. long and (10-x) cm. wide. What is the greatest volume possible? What will the dimensions of the box be?
I need all of the steps to get to the answer.
The volume as a function of x is given by:
V(x)=x(x+5)(10-x)=-x^3+5*x^2+50*x
To find the maximum volume, we differentiate V(x) with respect to x and equate the derivative to zero:
dV/dx = -3*x^2+10*x+50 =0
Solve for x to get:
x=-2.7 or x=6.08
We reject the negative value of x to retain
x=6.08.
To check if the Volume is a maximum, we calculate
V"(6.08)=d²V/dx²=10-6x=-26.48<0
so V(6.08) is a maximum.
To find the greatest volume possible for the rectangular box, we need to maximize the volume function V(x) in terms of x.
Step 1: Write the volume function.
The volume (V) of a rectangular box is determined by multiplying its length (L), width (W), and height (H). In this case, the length is given as (x+5) cm, the width is given as (10-x) cm, and the height is given as x cm. Therefore, the volume function can be written as:
V(x) = (x+5)(10-x)(x)
Step 2: Simplify the volume function.
Multiply the three terms together:
V(x) = x(x+5)(10-x)
Step 3: Expand the expression.
Use the distributive property to expand the expression:
V(x) = (10x - x^2 + 50x - 5x^2)
Step 4: Combine like terms.
Combine the like terms in the expression:
V(x) = 10x + 50x - x^2 - 5x^2
V(x) = -6x^2 + 60x
Step 5: Find the critical points.
To find the critical points, we need to find the values of x where the derivative of V(x) equals zero. Take the derivative of V(x) with respect to x:
V'(x) = -12x + 60
Set V'(x) = 0 and solve for x:
-12x + 60 = 0
-12x = -60
x = 5
Step 6: Check if the critical point is a maximum or minimum.
To determine if the critical point is a maximum or minimum, we need to examine the concavity of the function. Take the second derivative of V(x):
V''(x) = -12
Since V''(x) is negative for all values of x, it means that the critical point is a maximum.
Step 7: Plug the critical point into the volume function.
Evaluate the volume function at the critical point:
V(5) = -6(5)^2 + 60(5)
V(5) = -150 + 300
V(5) = 150
Therefore, the greatest volume possible for the rectangular box is 150 cubic centimeters. The dimensions of the box will be:
Height: x = 5 cm
Length: x + 5 = 5 + 5 = 10 cm
Width: 10 - x = 10 - 5 = 5 cm