a manufacturer wants to design an open box that has a square base and a surface area of 48sq units what dimension will produce a box with a maximum volume?
32
Let the side be x and the height h. Then x^2+4xh=48 and h=(48-x^2)/(4x). Now we have an expression for the volume in x: V(x)=x^2h=x^2*(48-x^2)/(4x)=12x-x^3/4. Taking the derivative we get V'(x)=12-3/4x^2=0. The volume is maximal when V'(x)=0 so 12-3/4x^2 and the positive solution is x=4. So h=(48-16)/16=2. We verify that is a maximum point by considering V''(x)=-3/2x so V''(2)=-3<0 and hence a mximum point. The dimension of the box is 4x4x2 units.
maths
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To find the dimensions that will produce a box with maximum volume, let's break down the problem step by step.
Step 1: Understand the Problem
We are given that the box has a square base, so all sides of the base are equal. We need to find the dimensions (length of the side of the square base and the height) that will result in the box having the maximum volume. The constraint provided is the surface area, which is 48 square units.
Step 2: Identify Variables
Let's assign variables to the dimensions of the box:
- Let s represent the side length of the square base.
- Let h represent the height of the box.
Step 3: Formulas
The surface area of the box can be determined by adding the areas of the individual sides:
Surface Area = area of the base + area of the four sides
= s^2 + 4sh
The volume of the box can be determined by multiplying the area of the base by the height:
Volume = s^2 * h
Step 4: Express One Variable in Terms of the Other
Since the surface area is given as 48 square units, we can express one variable in terms of the other using the surface area formula:
48 = s^2 + 4sh
Rearranging the equation, we get:
h = (48 - s^2) / (4s)
Step 5: Express the Volume Formula in Terms of One Variable
Substituting the expression for h in terms of s into the volume formula, we get:
Volume = s^2 * ((48 - s^2) / (4s))
Simplifying further, we have:
Volume = (1/4)(48s - s^3)
Step 6: Find the Critical Points
To find the dimensions that produce maximum volume, we need to find the critical points. These are the values of s where the derivative of the volume function is equal to zero.
Differentiating the volume function with respect to s, we get:
dV/ds = (1/4)(48 - 3s^2)
Setting dV/ds = 0 and solving for s, we find:
48 - 3s^2 = 0
3s^2 = 48
s^2 = 16
s = ±4
Since the side length of the box cannot be negative, we discard the negative solution. Therefore, s = 4.
Step 7: Find the Height
Substituting s = 4 into the expression for h, we find:
h = (48 - 4^2) / (4 * 4)
h = (48 - 16) / 16
h = 32 / 16
h = 2
Step 8: Answer
The dimensions that will produce a box with maximum volume are:
- Side length of the square base: 4 units
- Height of the box: 2 units