answer:Let the side length of the square base be x meters and the height of the box be h meters. The surface area of the box (without the open top) can be represented as: A = x^2 + 4xh Since we know that the total material used is 250 square meters, we can write: 250 = x^2 + 4xh Now, we need to find the dimensions that maximize the volume of the box. The volume of the box can be represented as: V = x^2h Our goal is to maximize V. To do this, we can first solve the surface area equation for h: h = (250 - x^2) / 4x Now, substitute this expression for h into the volume equation: V = x^2((250 - x^2) / 4x) Simplify the equation: V = (250x - x^3) / 4 To find the maximum volume, we can take the derivative of V with respect to x and set it equal to 0: dV/dx = (250 - 3x^2) / 4 = 0 Solve for x: 3x^2 = 250 x^2 = 250 / 3 x = √(250 / 3) x ≈ 9.082 Now, substitute the value of x back into the equation for h: h = (250 - (9.082)^2) / (4 * 9.082) h ≈ 6.055 So, the dimensions of the box that maximize its volume are approximately 9.082 meters for the side length of the square base and 6.055 meters for the height.

answer:Let the radius of the cylindrical container be r meters and the height be h meters. The volume V of the cylinder is given by the formula: V = πr^2h We are given that the volume is 1000 cubic meters, so: 1000 = πr^2h Now, we need to find the surface area of the cylinder, which will determine the cost of materials. The surface area A of a cylinder is given by the formula: A = 2πrh + 2πr^2 The cost of the materials for the side is 10 per square meter, and the cost for the top and bottom is 5 per square meter. So, the total cost C can be represented as: C = 10(2πrh) + 5(2πr^2) We want to minimize the cost C, so we need to find the dimensions r and h that will minimize C. To do this, we can use the method of Lagrange multipliers. We will introduce a new variable λ (the Lagrange multiplier) and form the following equation: ∇C = λ∇V Taking the gradient of C and V, we get: (∂C/∂r, ∂C/∂h) = λ(∂V/∂r, ∂V/∂h) Now, we can find the partial derivatives: ∂C/∂r = 20πh + 10πr ∂C/∂h = 20πr ∂V/∂r = 2πrh ∂V/∂h = πr^2 Now, we can set up the equations: 20πh + 10πr = λ(2πrh) 20πr = λ(πr^2) We also have the constraint equation from the volume: 1000 = πr^2h Now, we can solve the system of equations. From the second equation, we can find λ: λ = 20 Substituting λ back into the first equation, we get: 20πh + 10πr = 40πrh Divide both sides by 10πr: 2h + r = 4rh Now, we can use the constraint equation to eliminate h: h = 1000/(πr^2) Substitute this back into the equation above: 2(1000/(πr^2)) + r = 4r(1000/(πr^2)) Simplify the equation: 2000/(πr^2) + r = 4000/(πr) Now, cross-multiply and simplify: 2000πr = 4000r^3 - 4000r^2π Divide both sides by 2000r: π = 2r^2 - 2πr Now, solve for r: r^2 - πr + π/2 = 0 Using the quadratic formula, we find that the positive root is approximately: r ≈ 2.144 Now, we can find the height h using the constraint equation: h ≈ 1000/(π(2.144)^2) ≈ 2.144 So, the dimensions of the cylindrical container that minimize the cost of materials are approximately r = 2.144 meters and h = 2.144 meters.

answer:Let r be the radius of the base of the cylindrical can and h be its height. The volume of the can is given by the formula: V = πr^2h We are given that V = 200 cm³. So, 200 = πr^2h Now, we need to find the surface area of the can, which is the sum of the areas of the top, bottom, and side. The top and bottom are both circles with area πr^2, and the side is a rectangle that has been wrapped around the cylinder, with dimensions 2πr (the circumference of the base) and h (the height). So, the surface area A is given by: A = 2πr^2 + 2πrh The cost of materials is 0.08 per square centimeter for the top and bottom, and 0.05 per square centimeter for the side. So, the total cost C is given by: C = 0.08(2πr^2) + 0.05(2πrh) To minimize the cost, we need to minimize C with respect to r and h. We can first eliminate one variable by solving for h in terms of r using the volume equation: h = 200 / (πr^2) Now, substitute this expression for h into the cost equation: C = 0.16πr^2 + 0.1πr(200 / (πr^2)) Simplify the equation: C = 0.16πr^2 + 20 / r Now, we can find the minimum cost by taking the derivative of C with respect to r and setting it equal to 0: dC/dr = 0.32πr - 20 / r^2 = 0 To solve for r, we can multiply both sides by r^2: 0.32πr^3 - 20 = 0 Now, divide by 0.32π: r^3 = 20 / (0.32π) Take the cube root of both sides: r = (20 / (0.32π))^(1/3) r ≈ 1.216 cm Now, we can find the height h using the expression we found earlier: h = 200 / (π(1.216)^2) h ≈ 200 / (π(1.480)) h ≈ 43.019 cm So, the dimensions of the can that minimize the cost of materials are approximately r = 1.216 cm and h = 43.019 cm.

answer:Let the side length of the square base be x meters and the height of the container be h meters. The volume V of the container is given by: V = x^2 * h Since the volume is 1000 cubic meters, we have: 1000 = x^2 * h Now we need to find the surface area A of the container, which consists of the square base and four rectangular sides: A = x^2 + 4(xh) Our goal is to minimize the surface area A. To do this, we can first eliminate h from the surface area equation using the volume equation. From the volume equation, we have: h = 1000 / x^2 Substitute this expression for h into the surface area equation: A = x^2 + 4(x * (1000 / x^2)) Simplify the equation: A = x^2 + 4000 / x To minimize the surface area, we can find the critical points by taking the derivative of A with respect to x and setting it to zero: dA/dx = 2x - 4000 / x^2 Set dA/dx to zero and solve for x: 0 = 2x - 4000 / x^2 Multiply both sides by x^2: 0 = 2x^3 - 4000 Add 4000 to both sides: 2x^3 = 4000 Divide both sides by 2: x^3 = 2000 Take the cube root of both sides: x = (2000)^(1/3) ≈ 12.6 meters Now we can find the height h using the volume equation: h = 1000 / x^2 ≈ 1000 / (12.6^2) ≈ 6.3 meters So the dimensions of the container that will minimize its surface area are approximately 12.6 meters by 12.6 meters for the square base and 6.3 meters for the height.