Elementary Technical Mathematics

Published by Brooks Cole
ISBN 10: 1285199197
ISBN 13: 978-1-28519-919-1

Chapter 11 - Section 11.3 - Applications Involving Quadratic Equations - Exercise - Page 370: 11

Answer

(a) Dimensions = $5$ cm $\times$ $5$ cm (b) $7500$ cm$^3$

Work Step by Step

(a) Based on the illustration, we know the following: Length = $(60-2x)$ cm Width = $(40-2x)$ cm Area of the bottom of the container = $1500 cm^2$ We know that the formula for the area of a rectangle is Length $\times$ Width. We can use this formula to form a quadratic equation! Area = Length $\times$ Width $1500 = (40-2x)(60-2x)$ $1500=4x^2-200x+2400$ (Subtract 1500 from both sides to make the equation=0) $4x^2-200x+900=0$ We can then solve this equation by factoring! $4x^2-180x-20x+900=0$ $4x(x-45)-20(x-45)=0$ $(4x-20)(x-45)=0$ We then need to make each bracket = 0 to find the solutions for $x$. $(4x-20)=0$ or $(x-45)=0$ $x=5$ or $x=45$ We now have 2 solutions for $x$, but we can rule out 45 as if we substitute 45 into one of the dimensions, e.g: $60-2(45)$, we would get a length below 0 which is impossible. So, $x=5$ Dimensions = $5$cm $\times$ $5$ cm (b) For the volume, we need to figure out the length, width, and height. Length = $(60-2x)$ =$(60-2(5))$ =$50$ cm Width = $(40-2x)$ =$(40-2(5))$ =$30$ cm Height = $5$ cm (As this is the dimension of the squares cut out from each side) So, we can find the volume using the formula Volume = $lwh$ V = $lwh$ V = $(50)(30)(5)$ V = $7500$ cm$^3$
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