# Question 4, Exercise 5.3

Solutions of Question 4 of Exercise 5.3 of Unit 05: Polynomials. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

## Question 4

The volume of a rectangular solid is 2475 cubic units. The length of the box is three units more than twice the width of the box. The height is 2 units less than width. Find the dimensions of the box.

** Solution. **

Consider width = $x$ units

length = $2x+3$ units

height = $x-2$ units

Volume = 2475 cubic units.

By given condition \begin{align*} & x(2x+3)(x-2) = 2475 \\ \implies & x(2x^2+3x-4x-6)=2475 \\ \implies & x(2x^2-x-6)-2475=0 \\ \implies & 2x^3-x^2-6x-2475=0 \end{align*}

Suppose $$p(x)=2x^3-x^2-6x-2475.$$ Since \begin{align*} p(11)&=2(11)^3-11^2-6(11)-2475 \\ &=2662-121-66-2475 = 0 \end{align*} This gives $x=11$ is the zeros of $p(x)$. Thus

width = 11 units

length = $2(11)+3$ = 25 units

height = 11-2 = 9 units

Hence the dimension is 11 units by 25 units by 9 units.

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