a net enclosurefor practisinggolf shots is open at one end, as shown, find the dimensions that will minimize the amount of netting needed and give a volume of 144 m^3(netting is required only the sides, the top, the far end.)

Find 2 formulas that will use the dimensions, i.e. surface area and volume.

Let x be width and height (since has a square face)
Let y be length of the rect. prism

so you have 3 rectangles (bottom surface ignored) and a square at the far end.

S.A. (surface area) = w*h+ 3(l*w)
= x^2 + 3xy
=x(x+3y)
Vol. = l*w*h
144m^3 =x*x*y
rearrange vol. for y
144/x^2 = y
plug y into your S.A. equation and solve for x and simplify
Now find where the critical point (C.P.) of your S.A. formula using the 1st derivative of your new S.A. equation
S.A.' = 2x - 432/x^2
C.P. @ S.A.' = 0
0 = 2x - 432/x^2
x = 6
take your vol. equ that you isolated y and solve for y with your new x
y = 144/x^2
y = 144/(6)^2
y = 4
Thus the dimensions are 4*6*6 (L*W*H)

To minimize the amount of netting needed, we need to find the dimensions of the net enclosure that will give the minimum surface area.

Let's assume the dimensions of the net enclosure are as follows:
Length = L (in meters)
Width = W (in meters)
Height = H (in meters)

Since the net enclosure is open at one end, the volume is given as V = Length * Width * Height = 144 m^3.

We need to minimize the surface area, which consists of the sides, the top, and the far end. The area of the sides and top can be calculated as:

Area of sides = 2 * (Length * Height)
Area of top = Length * Width

Therefore, the total surface area is given by:

Surface area = Area of sides + Area of top + Area of far end
Surface area = 2 * (Length * Height) + Length * Width + Length * Height

Simplifying the equation, we get:

Surface area = 2 * (Length * Height) + Length * (Width + 2 * Height)

To find the dimensions that minimize the surface area, we need to differentiate the surface area equation with respect to Length, set it to zero, and solve for Length.

d(Surface area)/d(Length) = 2 * Height + Width + 2 * Height
0 = 2 * Height + Width + 2 * Height

Simplifying further, we have:

0 = 4 * Height + Width

Now, since we have two variables (Length and Width) and one equation, we need to eliminate one variable. Since we want to find the dimensions that minimize the netting needed, we can assume the width to be the smallest possible value, i.e., Width = 0.

Plugging in Width = 0 into the equation, we get:

0 = 4 * Height

Solving for Height, we find Height = 0.

However, the height cannot be zero as we need a three-dimensional enclosure. Therefore, the equation Width = 0 doesn't satisfy our requirements.

To find the actual dimensions that will minimize the netting needed, we need additional information or constraints, such as a fixed aspect ratio or specific range of values for the dimensions.

To find the dimensions that will minimize the amount of netting needed for the net enclosure, we need to use calculus and optimization techniques. Here's how you can find the solution step by step:

Step 1: Define the problem: The net enclosure is in the shape of a rectangular box open at one end. We want to find the dimensions that minimize the amount of netting used while maintaining a volume of 144 m³.

Step 2: Define the variables: Let's use x, y, and z to represent the dimensions of the box.

Step 3: Set up the equations: We have the following constraints:
- The volume of the box is V = x * y * z = 144 m³.
- The surface area of the box is A = xy + 2xz + 2yz.

Step 4: Solve for one variable in terms of the other two: From the volume equation, we can solve for one variable in terms of the other two. Let's solve for z:
z = 144 / (xy).

Step 5: Substitute the expression for z into the surface area equation: Replace z in the surface area equation with the expression we found in step 4:
A = xy + 2x(144 / (xy)) + 2y(144 / (xy)).
Simplify this to: A = xy + (288 / y) + (288 / x).

Step 6: Minimize the surface area equation: To find the minimal surface area, we need to find the critical points of the surface area equation. Take the derivatives of A with respect to x and y separately and set them equal to zero:
dA/dx = y - (288 / x²) = 0,
dA/dy = x - (288 / y²) = 0.

Step 7: Solve the derivative equations simultaneously: Solve the derivative equations to find the critical points. Solving these equations will give us the values of x and y that minimize the surface area.

Step 8: Substitute the values of x and y into the volume equation to find z: Use the x and y values found in step 7 to find the corresponding value of z from the volume equation.

Step 9: Check if the solution is valid: Make sure that the resulting dimensions obtained from step 8 give a valid box shape (positive dimensions) and meet the given constraints.

By following these steps, you will be able to find the dimensions (x, y, and z) that will minimize the amount of netting needed for the given net enclosure with a volume of 144 m³.

Let length be L and width of open end be W. Height is then 144/(LW)

Area of netting required
A = LW (top) + WH (end) + 2LH (sides)
A = LW + 144/L + 288/W

Differentiate to find dA/dL and dA/dW and equate both to zero. Solve these as a pair of simultaneous equations. You should end up with L = 4.16m, W = 8.32m, H = 4.16m (2 dec. pl.)