Question

Ray and Heather are upgrading the storage shed in their backyard. The existing shed has

the shape of a rectangular prism with dimensions 1 m by 2 m by 4 m. By what amount

should they increase each dimension so that the new shed is 9 times the volume of the

original shed? Provide a full algebraic solution.

Answer 1

Define the following for the dimensions of the rectangular prism.

$w=width, \space l=length,\space h=height$

The volume of the original prism is given as,

$V_1=l_1\times w_1 \times h_1=1\times2\times 4=8m^3$

The volume of the new rectangular prism should be 9 times the volume of the original rectangular prism. Therefore, volume of the new rectangular prism is,

$V_2=V_1\times 9=8\times9=72.$

Let $a$ be the increase made for each dimension. The dimensions for the new rectangular prism are given as,

$l_2=l_1+a=1+a$

$w_2=w_1+a=2+a$

$h_2=h_1+a=4+a$

The volume of the new rectangular prism is,

$V_2=l_2\times w_2\times h_2=(1+a)\times(2+a)\times(4+a)=72......(i)$

Opening the brackets for equation $(i)$ above gives,

$a^3+7a^2+14a+8=72...........(ii)$

We shall solve for the value of $a$ in equation $(ii)$ as described below.

Let $a=0$ then equation $(ii)$ gives,

$(0)^3+7(0)^2+14(0)+8=8\not=72$. Thus , when $a=0$ equation $(ii)$ does not hold.

Let $a=1$ then equation $(ii)$ gives,

$(1)^3+7(1)^2+14(1)+8=30\not=72$. Thus , when $a=1$ equation $(ii)$ does not hold.

Let $a=2$ then equation $(ii)$ gives,

$(2)^3+7(2)^2+14(2)+8=8+28+28+8=72$. Thus, when $a=2$ equation $(ii)$ holds.

Therefore, each dimension should be increased by $a=2\space m$ so that the new shed is 9 times the volume of the original shed.