# Calculus grade 12: Optimisation Practice

TLDRIn this instructional video, the presenter tackles an optimization problem involving a rectangular box with given volume and material costs. The challenge is to minimize construction costs by determining the box's dimensions. The solution process includes deriving the box's surface area formula, substituting the volume equation to eliminate one variable, and applying calculus to find the minimum cost. The presenter skillfully guides viewers through the mathematical steps, emphasizing the importance of understanding shapes and formulas, and concludes with the optimal box width calculation.

###### Takeaways

- π The problem involves optimizing the dimensions of a rectangular box with given volume and material costs for its construction.
- π The box has a fixed volume of 5 cubic meters, with the length being three times the width (l = 3w).
- π‘ The volume formula is used to express the height in terms of width and volume: h = 5 / (3w^2).
- π’ The surface area of the box is calculated by considering the areas of the top, bottom, and sides, resulting in the formula 2wh + 6w^2 + 6wh.
- πΈ The cost of constructing the box is determined by the surface area and the cost per square meter for the top/bottom and sides, leading to the cost formula 48wh * 6 + 6w^2 * 15.
- π To minimize the cost, the first derivative of the cost function with respect to width (w) is taken and set to zero.
- π The derivative involves simplifying the cost function to a single variable expression, 80/w + 90w, before differentiation.
- βοΈ The first derivative of the simplified cost function is calculated as -80/w^2 + 180w.
- π The minimum cost condition is found by setting the derivative equal to zero and solving for width (w).
- π The solution to the derivative equation yields the width of the box as approximately 0.76 meters.
- π The final answer is given in meters, consistent with the units used for volume and surface area in the problem.

###### Q & A

### What are the initial dimensions of the rectangular box provided in the problem?

-The problem involves a rectangular box with dimensions length (L), width (W), and height (H). The length of the base is given as three times the width (L = 3W).

### How is the volume of the rectangular box related to its dimensions?

-The volume (V) of the box is calculated as the product of its length, width, and height, i.e., V = L Γ W Γ H. The volume is given as 5 cubic meters, so 5 = 3W Γ W Γ H.

### What is the formula for the surface area of the rectangular box?

-The surface area (SA) of the rectangular box is derived by summing the areas of all its faces: SA = 2(WH) + 2(3W^2) + 2(3WH). This simplifies to SA = 8WH + 6W^2.

### How is the cost to construct the box determined?

-The cost is determined by multiplying the surface area of the top and bottom by 15 (as they cost 15 rand per square meter) and the surface area of the sides by 6 (as they cost 6 rand per square meter). The final cost expression is Cost = 48WH + 90W^2.

### What is the relationship between the surface area and cost of the box?

-The cost function is directly related to the surface area of the box. It combines the costs of the top, bottom, and sides of the box based on their respective surface areas and the cost per square meter of the material used.

### What is the significance of finding the width that minimizes the cost?

-Finding the width that minimizes the cost involves optimizing the cost function, which is crucial in ensuring the construction of the box is as economical as possible.

### Why is it necessary to express the height (H) in terms of the width (W) before differentiating the cost function?

-It's necessary to express the height (H) in terms of width (W) because the cost function initially involves two variables (W and H). For optimization (finding minimum or maximum), the cost function must be expressed in terms of a single variable.

### How is the first derivative of the cost function used to find the minimum cost?

-The first derivative of the cost function with respect to W is set to zero to find the critical points, which correspond to either a minimum or maximum cost. The specific point where the derivative equals zero gives the width (W) that minimizes the cost.

### What is the final expression for the cost function after substituting H in terms of W?

-After substituting H in terms of W using the volume equation, the cost function becomes C(W) = 80/W + 90W^2.

### What is the width (W) that minimizes the cost, based on the final calculation?

-The width that minimizes the cost is approximately 0.76 meters, as determined by solving the equation derived from the first derivative of the cost function.

###### Outlines

##### π Problem Introduction and Box Dimensions

The video script begins with the introduction of an optimization problem involving a rectangular box. The presenter attempts to draw the box and explains the dimensions: length (l), width (w), and height (h). It's specified that the length is three times the width (l = 3w). The volume of the box is given as a constant, five cubic units. The focus then shifts to the cost of constructing the box, with the material for the top and bottom costing 15 units per square meter, and the sides costing six units per square meter. The presenter emphasizes the importance of calculating surface area, not volume, for determining the cost.

##### π Calculating Surface Area and Cost

The presenter proceeds to calculate the surface area of the box by identifying and summing the areas of its different faces. The front and back faces are each w times h, with two such faces, resulting in 2wh. The top and bottom, each being 3w times w (since l = 3w), contribute 6w squared to the total area. The sides are h times 3w each, with two sides, adding 6wh to the area. The presenter then combines like terms to express the total surface area as 8wh + 6w squared. The cost is calculated by multiplying the surface area by the respective material costs for the top/bottom and sides, resulting in a cost formula involving w and h.

##### π Optimization for Minimum Cost

The script's focus turns to optimization, aiming to minimize the cost of constructing the box. The presenter recalls the volume formula and uses it to express the height (h) in terms of w, which is then substituted into the cost formula. This substitution simplifies the cost formula to a function of w alone, allowing for optimization. The presenter explains the process of finding the minimum cost by taking the first derivative of the cost function with respect to w and setting it to zero. After simplifying, the derivative leads to a cubic equation in w. Solving this equation gives the width that minimizes the cost. The presenter concludes by calculating the width to be approximately 0.76 meters, ensuring the units are consistent with the problem's context.

###### Mindmap

###### Keywords

##### π‘Rectangular Box

##### π‘Dimensions

##### π‘Volume

##### π‘Optimization

##### π‘Surface Area

##### π‘Cost

##### π‘First Derivative

##### π‘Minimum

##### π‘Graph

##### π‘Cube Root

##### π‘Meters

###### Highlights

Introduction to the optimization problem involving a rectangular box with given dimensions.

The base length is three times the width, represented as "l = 3w".

The volume of the box is given as a constant value of 5 cubic meters.

Volume formula is expressed as "V = l Γ w Γ h".

Cost of materials for the top and bottom is 15 rand per square meter.

Material for the sides of the box costs 6 rand per square meter.

The cost function for constructing the box is introduced.

Surface area of the box is calculated by identifying and adding the areas of its different shapes.

Total surface area is expressed as "2wh + 6w^2 + 6wh".

The cost is calculated by multiplying the surface area components with their respective material costs.

The cost function is simplified to "C = 48wh + 90w^2".

The volume equation is used to express the height "h" in terms of width "w".

The cost function is rewritten with "h = 5 / 3w^2" substituted for "h".

The cost function is simplified further to "C = 80 / w + 90w".

The process of finding the minimum cost involves taking the first derivative of the cost function.

The first derivative of the cost function is calculated as "dc/dw = -80/w^2 + 180w".

Setting the first derivative equal to zero to find the minimum cost.

Solving for "w" gives the width that minimizes the cost, "w = 4/3" meters.

The final answer is presented with the width in meters, ensuring the units are consistent with the problem's context.

###### Transcripts

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