an open box with dimension 2 inches x 3 inches x 4 inches needs to be increase in size to hold 5 times as much as material as the current box?
a. write a function that represents the new box
b.find the dimension of the new box
c.make a mathemathical solution in getting the new dimension
Volume=w*l*h and the new box
volume=5*w*l*h
so solutions can be new w=5*oldw=10
10x3x4 is a solution, as is
2*15*4 is a solution, as it
4*3*10 is a solution
Thank you for your help
yes Ms.Sue you are great and helpful
To find the dimensions of the new box that can hold 5 times as much material as the current box, you can follow these steps:
a. Write a function that represents the new box:
Let's denote the dimensions of the current box as length (L), width (W), and height (H). The volume of the current box (V_current) can be calculated using the formula:
V_current = L * W * H
To find the dimensions of the new box, we need to find the new values of length (L_new), width (W_new), and height (H_new) that will result in a volume five times greater than the volume of the current box.
b. Find the dimension of the new box:
The volume of the new box (V_new) should be equal to 5 times the volume of the current box (V_current):
V_new = 5 * V_current
c. Make a mathematical solution to get the new dimensions:
1. Start by calculating the volume of the current box (V_current) using the given dimensions: L = 2 inches, W = 3 inches, H = 4 inches:
V_current = L * W * H = 2 * 3 * 4 = 24 cubic inches
2. Calculate the volume of the new box (V_new) by multiplying V_current by 5:
V_new = 5 * V_current = 5 * 24 = 120 cubic inches
3. Solve for the new dimensions using the volume formula:
V_new = L_new * W_new * H_new
Since we need to find three unknowns (L_new, W_new, H_new), we can rearrange the equation:
L_new * W_new * H_new = V_new
It is important to note that multiple combinations of dimensions could yield the same volume. For simplicity, we will assume that all dimensions are increased by the same factor (denoted as x).
Thus, we can write the equations:
L_new = x * L
W_new = x * W
H_new = x * H
Substituting these into the volume equation, we have:
x * L * x * W * x * H = V_new
x^3 * L * W * H = V_new
Substituting the known values and solving for x:
x^3 * 2 * 3 * 4 = 120
24x^3 = 120
x^3 = 120 / 24
x^3 = 5
x = ∛5
Therefore, the new dimensions of the box will be:
L_new = ∛5 * L = ∛5 * 2 inches
W_new = ∛5 * W = ∛5 * 3 inches
H_new = ∛5 * H = ∛5 * 4 inches