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.

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

Further explanation:

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

Consider the length of printed area is x and width of the printed area is [tex]y[/tex] .

Let [tex]A[/tex] 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 [tex]A = xy[/tex].

Here, area is given as [tex]210{\text{ i}}{{\text{n}}^2}[/tex] .

Substitute [tex]210{\text{ i}}{{\text{n}}^2}[/tex] for [tex]A[/tex] in equation [tex]A = xy[/tex] as follows:

[tex]\begin{aligned}210&=xy\\y&=\frac{{210}}{x}{\text{}}\\\end{aligned}[/tex] ......(1)

From Figure 1, the length of inner rectangle is [tex]x - 2[/tex] and width is [tex]y - 3[/tex].

Area is calculated as follows:

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

Substitute [tex]\frac{{210}}{x}[/tex] for [tex]y[/tex] in above equation.

[tex]\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}[/tex] ......(2)

Derivate the equation (2) with respect to [tex]x[/tex] as follows:

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

Substitute [tex]0[/tex] for [tex]A'[/tex] in equation (3) to obtain the value of

[tex]\begin{aligned}0&= - 3+\frac{{420}}{{{x^2}}}\\3&=\frac{{420}}{{{x^2}}}\\{x^2}&= \frac{{420}}{3}\\{x^2}&=140\\\end{aligned}[/tex].

Further simplify the above equation.

[tex]\begin{aligned}x&=\pm\sqrt{140}\\&=\pm\sqrt{2\cdot2\cdot35}\\&=\pm\,2\sqrt {35}\\\end{aligned}[/tex]

Therefore, the value of [tex]x[/tex] is [tex]\,2\sqrt{35}\,{\text{ or }}-2\sqrt{35}[/tex].

Derivate the equation (3) as follows.

[tex]\begin{aligned}A''&=0-\frac{{420}}{{{x^3}}}\\&=0-\frac{{420}}{{{x^3}}}\\\end{aligned}[/tex]

Now, if the value of [tex]x[/tex] is positive, then [tex]A''[/tex] must be negative and vice versa.

So, for obtaining the maximum dimension, the value of [tex]x[/tex] must be positive.

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

[tex]\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}[/tex]

Therefore, the dimensions are [tex]\,{\mathbf{2}}\sqrt{{\mathbf{35}}}\,\,{\mathbf{ \times }}\,\,{\mathbf{3}}\sqrt {{\mathbf{35}}}{\mathbf{ inches}}[/tex].

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

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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