Year 10 Maths Find the length and width that will maximize the area of rectangle

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SUMMARY

The discussion focuses on maximizing the area of a rectangle given the constraint equation \(5W + 2L = 550\). The area \(A\) is expressed as \(A = LW\), and through substitution, the equation transforms into \(A = 110L - \frac{2L^2}{5}\). The maximum area occurs at the vertex of this quadratic equation, specifically when \(L = \frac{275}{2}\). Calculating the corresponding width \(W\) and area \(A\) from this value is the next logical step.

PREREQUISITES
  • Understanding of algebraic manipulation and quadratic equations
  • Familiarity with the concept of maximizing functions
  • Knowledge of the vertex form of a parabola
  • Basic understanding of geometry related to rectangles
NEXT STEPS
  • Calculate the width \(W\) using \(W = \frac{A}{L}\) after determining \(L\)
  • Explore the implications of changing the constraint \(5W + 2L = 550\) on the area
  • Investigate the use of calculus to find maximum values of functions
  • Learn about optimization problems in geometry and their applications
USEFUL FOR

Students studying Year 10 mathematics, educators teaching optimization in geometry, and anyone interested in applying algebra to real-world problems.

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View attachment 9234

The question is in the image. Working out with every step would be much appreciated.
 

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Here's a start:

Let $W$ be width, $L$ be length an $A$ be the desired area. Then,

$$5W+2L=550$$

$$LW=A$$

Can you make any progress from there?
 
Greg said:
Here's a start:

Let $W$ be width, $L$ be length an $A$ be the desired area. Then,

$$5W+2L=550$$

$$LW=A$$

Can you make any progress from there?

$$W=\frac AL$$

$$\frac{5A}{L}+2L=550$$

$$5A+2L^2=550L$$

$$A=110L-\frac{2L^2}{5}$$

$A$ has a maximum at the vertex of this inverted parabola, so $L=\frac{275}{2}$. Finding $A$ and $W$ from here should be straightforward.
 

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