Question

A poster is to have an area of 210 in2 with 1 inch margins at the bottom and sides and a 2 inch margin at the top. find the exact dimensions that will give the largest printed area.

Answer 1

Let

x-------> the length side of the poster

y-------> the width side of the poster

we know that

Area of the poster is equal to

$210=x*y\\\\ y=\frac{210}{x}$ -------> equation $1$

Area of the printed area is equal to

$Aprinted=(x-2)*(y-3)\\ Aprinted=xy-3x-2y+6$ ----> equation $2$

substitute equation $1$ in equation $2$

$Aprinted=[210]-3x-2*[\frac{210}{x}]+6$

using a graph tool

Find the vertex of the graph

The vertex of the graph is the point that represent the value of x for the largest printed area

see the attached figure

the vertex is the point $(11.83,145)$

the largest printed area is $145 in^{2}$

$x=11.83 in$

find the value of y

$y=\frac{210}{x} \\ \\ y=\frac{210}{11.83} \\ \\ y=17.75in$

therefore

the answer is

The dimensions of the poster are $11.83 in x 17.75 in$

Answer 2

Theexact dimensions that will give the largest printed area is $\boxed{\,{\mathbf{2}}\sqrt {{\mathbf{35}}} \,\,{\mathbf{ \times }}\,\,{\mathbf{3}}\sqrt {{\mathbf{35}}} {\mathbf{ inches}}}$.

Further explanation:

It is given that the area of the poster is $210{\text{ i}}{{\text{n}}^2}$  with $1{\text{ inch}}$ at the bottom and sides and $2{\text{ inch}}$ margin at the top.

Consider the length of printed area is x  and width of the printed area is $y$ .

Let $A$ be the printed area.

The dimension of poster and area is shown below in Figure 1.

Now, area is the multiplication of length and width that is $A = xy$.

Here, area is given as $210{\text{ i}}{{\text{n}}^2}$ .

Substitute $210{\text{ i}}{{\text{n}}^2}$ for $A$ in equation $A = xy$ as follows:

$\begin{aligned}210&=xy\\y&=\frac{{210}}{x}{\text{}}\\\end{aligned}$      ......(1)

From Figure 1, the length of inner rectangle is $x - 2$ and width is $y - 3$.

Area is calculated as follows:

$\begin{aligned}A&=\left({{\text{length}}}\right)\cdot\left({{\text{width}}}\right)\\&= \left({x - 2}\right)\cdot\left({y - 3}\right)\\\end{aligned}$

Substitute $\frac{{210}}{x}$ for $y$ in above equation.

$\begin{aligned}A&=\left({x - 2}\right)\left({\frac{{210}}{x} - 3}\right)\\&=210 - 3x-\frac{{420}}{x}+6\\&=216 - 3x - \frac{{420}}{x}{\text{ }}\\\end{aligned}$      ......(2)

Derivate the equation (2) with respect to $x$ as follows:

$\begin{aligned}A'&= 0 - 3 + \frac{{420}}{{{x^2}}}\\A'&= - 3+\frac{{420}}{{{x^2}}}{\text{}}\\\end{aligned}$  ......(3)

Substitute $0$ for $A'$ in equation (3) to obtain the value of  

$\begin{aligned}0&= - 3+\frac{{420}}{{{x^2}}}\\3&=\frac{{420}}{{{x^2}}}\\{x^2}&= \frac{{420}}{3}\\{x^2}&=140\\\end{aligned}$.

Further simplify the above equation.

$\begin{aligned}x&=\pm\sqrt{140}\\&=\pm\sqrt{2\cdot2\cdot35}\\&=\pm\,2\sqrt {35}\\\end{aligned}$

Therefore, the value of $x$ is $\,2\sqrt{35}\,{\text{ or }}-2\sqrt{35}$.

Derivate the equation (3) as follows.

$\begin{aligned}A''&=0-\frac{{420}}{{{x^3}}}\\&=0-\frac{{420}}{{{x^3}}}\\\end{aligned}$

Now, if the value of $x$ is positive, then $A''$ must be negative and vice versa.

So, for obtaining the maximum dimension, the value of $x$ must be positive.

Substitute $\,2\sqrt{35}$ for $x$ in equation (1) to obtain the value of $y$.

$\begin{aligned}y&=\frac{{210}}{{\,2\sqrt{35}}}\\&=\frac{{105}}{{\,\sqrt {35}}}\\&=\frac{{105}}{{\sqrt {35}}}\left({\frac{{\sqrt{35}}}{{\sqrt{35}}}}\right)\\&={\mathbf{3}}\sqrt{{\mathbf{35}}}\\\end{aligned}$

Therefore, the dimensions are $\,{\mathbf{2}}\sqrt{{\mathbf{35}}}\,\,{\mathbf{ \times }}\,\,{\mathbf{3}}\sqrt {{\mathbf{35}}}{\mathbf{ inches}}$.

Thus, theexact dimensions that will give the largest printed area is $\boxed{\,{\mathbf{2}}\sqrt {{\mathbf{35}}}\,\,{\mathbf{ \times }}\,\,{\mathbf{3}}\sqrt {{\mathbf{35}}}{\mathbf{ inches}}}$.

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Answer Details:

Grade: Junior High School

Subject: Mathematics

Chapter: Surface Area and Volumes

Keywords: Surface area, linear equation, system of linear equations in two variables, largest printed area