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
