a closed box has a fixed surface area A and a suqare base wide side X
a) find the formula for its volume, v, as a function x
b) sketch the graph of v against x
c) find the max value of V
is that the answer for a ?
Volume of the box is x^2h and if you plug in the value of h, you get,
V=1/4(Ax-2x^3)/x
To find the formula for the volume of a closed box with a fixed surface area A and a square base of side X, we can start by visualizing the box and breaking it down into its component parts.
Let's assume that the box has height H and the four sides of the box are identical rectangles.
a) Finding the formula for the volume (V) as a function of X:
The area of the base of the box is X^2 (since it is a square base).
The area of the four sides of the box, excluding the base, is given by the formula 2XH (since the width of each side is X and the height is H). Since there are four identical sides, the total area of the four sides is 4XH.
The total surface area (A) of the box is the sum of the area of the base and the area of the four sides: A = X^2 + 4XH.
Now, we need to express the height (H) in terms of X. To do this, we can rearrange the equation for the surface area to solve for H:
4XH = A - X^2
H = (A - X^2) / (4X)
Finally, we can substitute this expression for H into the formula for volume:
V = X^2 * H = X^2 * ((A - X^2) / (4X))
Simplifying further:
V = (A * X^2 - X^4) / (4X)
V = (A * X - X^3) / 4
Therefore, the formula for the volume of the closed box is V = (A * X - X^3) / 4.
b) Sketching the graph of V against X:
To sketch the graph, plot values of X on the x-axis and corresponding values of V on the y-axis using the formula V = (A * X - X^3) / 4. Choose a suitable range of X values and calculate the corresponding V values. Then, plot the points and connect them to visualize the graph.
c) Finding the maximum value of V:
To find the maximum value of V, we can take the derivative of the volume function with respect to X and set it equal to zero.
dV/dX = (A - 3X^2) / 4
Setting dV/dX equal to zero:
(A - 3X^2) / 4 = 0
A - 3X^2 = 0
3X^2 = A
X^2 = A / 3
X = sqrt(A / 3)
Therefore, the maximum value of V occurs when X = sqrt(A / 3).
area= 2x^2+4xh
Volume= hx^2
h= (area-2x^2)/4x
volume= 1/4 (area-2x^2)x