D AY 147 – S QUARE AND RECTANGULAR DESIGNS
I NTRODUCTION Geometry can be used in modeling of wonderful designs like designs of rectangular house roofs, a rectangular table top, a rectangular field, a rectangular window frame, among other designs. In order to design these objects, we aim at minimizing the materials used to reduce the cost of design due to the scarcity of necessary materials. In such a case, we use geometric methods and algebra to come up with the optimum solution that will best utilize the available materials and at the same time minimize the cost of design. In this lesson, we will learn how to apply geometric methods to solve design and structural problems involving rectangles and squares to satisfy physical constraints or to minimize cost.
V OCABULARY Square A quadrilateral with all sides congruent and opposite sides parallel. It has four right angles. Rectangle A quadrilateral whose opposite sides are congruent and parallel. It has four right angles. Physical constraint A limit or restriction in supply or availability of a physical quantity or a natural property.
A number of steps are required in solving design problems. The steps should be carried out in the correct order in order to achieve the required objectives. Knowledge of basic shapes and formulae that describe these shapes is important in solving design problems. Forming and solving algebraic expressions and equations is equally important. The geometric concepts learned help in identification of the appropriate formula to best serve the objectives of the problem.
S OLVING DESIGN PROBLEMS INVOLVING SQUARES AND RECTANGLES Understand the objectives of the design problem. Draw a sketch or diagram to illustrate the problem. Identify the variables used and label the sketch appropriately using these variables. Identify the equation that illustrates the objective of the problem. Identify the equation that illustrates any constraints. Express the objective equation in terms of a single variable by the help of the equation illustrating the constraints.
Write down all the corresponding expressions or equations. Apply algebraic methods to solve the problem. Verify your solution.
Example 1 What is the largest possible rectangular area you can enclose with 256 feet of fencing wire? Solution The objective of the problem is to find the largest possible area that can be enclosed by a 256 feet wire. l w We can sketch the rectangular area as shown above.
The perimeter equation will be given as: 2 𝑚 + 𝑥 = 256 This is the constraint equation . 𝑚 represents the length and 𝑥 represents the width. This simplifies to: 𝑚 + 𝑥 = 128 … . 1 We then write the area equation as: 𝐵 = 𝑚𝑥 … (2) This is the objective equation . Equation 1 can be rewritten as: 𝑚 = 128 − 𝑥 Substituting this in equation 2, we have: 𝐵 = 128 − 𝑥 𝑥 𝐵 = 128𝑥 − 𝑥 2
We rewrite the equation in vertex form where the coordinates 𝑊 ℎ, 𝑙 are the coordinates of the vertex of the parabola: The objective equation can be rewritten as: 𝐵(𝑥) = −𝑥 2 + 128𝑥 We solve by completing square 𝐵 𝑥 = − 𝑥 2 − 128𝑥 = −(𝑥 2 − 128𝑥 + 64 2 − 64 2 ) = − 𝑥 2 − 128𝑥 + 64 2 + 64 2 − 𝑥 − 64 2 + 64 2 = − 𝑥 − 64 + 4096 Since the coefficient of 𝑥 2 is negative, the parabola faces down hence, it has a maxima. The maxima is at a point ℎ, 𝑙 = (64,4096) This shows that the largest possible rectangular area will be 𝟓𝟏𝟘𝟕 square feet .
Example 2 A kindergarten teacher has a 500 feet rope which she intends to use to form a rectangular playground having three equal and parallel sections for three groups of pupils as illustrated below. (a) Find the maximum possible area of the playground. (b) The dimensions that will give the maximum area. l w
Solution The objective is to maximize the area of the play area. The play area can be labelled as shown below. l w w w w l
The constraint equation becomes: 2𝑚 + 4𝑥 = 500 The objective equation becomes: 𝐵 = 𝑚𝑥 Expressing the objective equation in terms of one variable, say, 𝑥 we have: From 2𝑚 + 4𝑥 = 500 ⟹ 𝑚 = 250 − 2𝑥 𝐵 = 250 − 2𝑥 𝑥 = 250𝑥 − 2𝑥 2 𝐵 = 250𝑥 − 2𝑥 2
We solve the equation by completing square 𝐵(𝑥) = −2𝑥 2 + 250𝑥 = −2 𝑥 2 − 125𝑥 = −2 𝑥 2 − 125𝑥 + 62.5 2 − 62.5 2 = −2 𝑥 2 − 125𝑥 + 62.5 2 + 7812.5 = −2 𝑥 − 62.5 2 + 7812.5 The leading coefficient is negative hence, the function is facing down. This implies that it has a maxima. This occurs at (62.5,7812.5) This shows that the maximum area will be 𝟖𝟗𝟐𝟑. 𝟔 square feet .
(b) We have found out that 𝑥 = 62.5 then the length will be: 𝑚 = 𝑞 − 2𝑥 𝑚 = 250 − 2 62.5 = 125 The dimensions will therefore be 125 ft. by 62.5 ft.
HOMEWORK Arnold wants to enclose a rectangular region of his farm for growing flowers using a 1000 feet of fencing material. The region he wants to enclose is bounded by a wall fence on one side and therefore will not require fencing. Which dimensions of the region will enable him to enclose the largest area?
A NSWERS TO HOMEWORK 500 ft. by 250 ft.
THE END
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