Geometric Design Problems
Aligned to G-MG.A.3 — Common Core State Standards for Mathematics.
What this lesson teaches
Design problems ask you to choose dimensions or shapes to meet constraints such as minimizing cost, fitting a space, or maximizing volume. Set up an equation from the geometry, then solve for the value that satisfies the requirement.
Worked examples
Example 1
Design an open-top box from a 12-by-12 cm sheet by cutting squares of side x from each corner and folding. The base becomes (12 - 2x) by (12 - 2x) with height x, so volume V = x(12 - 2x)^2. Test x = 2: V = 2(12 - 4)^2 = 2(8)^2 = 2(64) = 128 cubic cm. Trying values shows x = 2 gives the maximum whole-number volume for this sheet.
Example 2
Design the cheapest cylindrical can holding 355 cubic centimeters. The constraint is pi*r²*h = 355, so h = 355/(pi*r²). The cost follows the surface area S = 2*pi*r² + 2*pi*r*h = 2*pi*r² + 710/r. Test values: r = 3 gives S = 56.5 + 236.7 = 293 cm²; r = 4 gives S = 100.5 + 177.5 = 278 cm²; r = 5 gives S = 157 + 142 = 299 cm². The minimum sits near r = 4 cm.
Example 3
Common mistake: optimizing without writing the constraint. Asked for the box of largest volume from a 12-by-12 sheet, a student may just try making x large — but x also shrinks the base, and at x = 6 the base vanishes and the volume is 0. Every design problem has a constraint that ties the variables together; write it down, use it to reduce to ONE variable, and only then hunt for the best value.
Practice questions
- A rectangular pen must have area 48 square meters with one side 6 m. Find the other side.
- You have 20 m of fencing for a rectangular garden. Give the dimensions of a square that uses it all, and its area.
- A cylindrical can must hold about 500 cubic cm. If its radius is 5 cm, find the required height (use pi about 3.14).
Watch the lesson
Every lesson comes with a video taught in English and Spanish — the same video the QR code in the printed workbook opens.
▶ Watch this lessonEn español
Problemas de diseño geométrico
Los problemas de diseño piden elegir dimensiones o formas para cumplir restricciones como minimizar costo, caber en un espacio o maximizar el volumen. Plantea una ecuación a partir de la geometría, luego resuelve el valor que satisface el requisito.
Ejemplo 1: Diseña una caja sin tapa con una hoja de 12 por 12 cm cortando cuadrados de lado x en cada esquina y doblando. La base queda (12 - 2x) por (12 - 2x) con altura x, así que el volumen V = x(12 - 2x)^2. Prueba x = 2: V = 2(12 - 4)^2 = 2(8)^2 = 2(64) = 128 cm cúbicos. Al probar valores se ve que x = 2 da el volumen máximo en números enteros para esta hoja.
Ejemplo 2: Diseña la lata cilíndrica más barata que contenga 355 centímetros cúbicos. La restricción es pi*r²*h = 355, así que h = 355/(pi*r²). El costo sigue al área superficial S = 2*pi*r² + 2*pi*r*h = 2*pi*r² + 710/r. Prueba valores: r = 3 da S = 56.5 + 236.7 = 293 cm²; r = 4 da S = 100.5 + 177.5 = 278 cm²; r = 5 da S = 157 + 142 = 299 cm². El mínimo está cerca de r = 4 cm.
Ejemplo 3: Error común: optimizar sin escribir la restricción. Al pedir la caja de mayor volumen a partir de una hoja de 12 por 12, alguien puede simplemente intentar agrandar x — pero x también encoge la base, y en x = 6 la base desaparece y el volumen es 0. Todo problema de diseño tiene una restricción que ata las variables; escríbela, úsala para reducir a UNA variable y solo entonces busca el mejor valor.
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