A pixel is the smallest unit of a digital image.
The number of pixels/cm in each of the horizontal and vertical directions of a digital image affects the quality of the image. The more pixels/cm, the sharper the image will be.
A small monitor has dimensions \(15\) cm by \(10\) cm and has \(80\) pixels/cm in each dimension. The total number of pixels is \((15\times 80)\times (10\times 80)= 960\,000\).
The manufacturer wants to build a new monitor with \(2\,145\,624\) pixels. To accomplish this, both the length and width of the screen will be increased by \(n\%\) and the number of pixels/cm in each dimension will be increased by \(2n\%\). Determine the dimensions of the new monitor and the new number of pixels/cm.
Let \(a>0\) represent the percentage increase, expressed as a decimal, in each dimension of the monitor. Then \(2a\) represents the percentage increase, expressed as a decimal, in the number of pixels/cm. (So \(n=100a\) and \(2n=200a\).)
We know that \[[(\text{New Length}) \times (\text{New pixels/cm})] \times [(\text{New Width}) \times (\text{New pixels/cm})]\] \[= \text{Total Number of Pixels}\] Thus, \[\begin{align*} [15(1+a) \times 80(1+2a)] \times [10(1+a) \times 80(1+2a)] &= 2\,145\,624\\ (15)(80)(10)(80)(1+a)^2(1+2a)^2&=2\,145\,624\\ 960\,000(1+a)^2(1+2a)^2&=2\,145\,624\\ (1+a)^2(1+2a)^2&=2.235\,025 \end{align*}\] Taking the square root of both sides, we have \((1+a)(1+2a)=\pm 1.495\).
Since \(a>0\), we have \((1+a)(1+2a)>0\). Thus, \((1+a)(1+2a)=1.495\).
Expanding and simplifying, we have \(2a^2+3a+1=1.495\), and so \(2a^2+3a-0.495=0\).
Using the quadratic formula, \(a=\dfrac{-3\pm\sqrt{9-4(2)(-0.495)}}{4}=\dfrac{-3\pm\sqrt{12.96}}{4}=\dfrac{-3\pm 3.6}{4}\).
It follows that \(a=0.15\) or \(a=-1.65\). Since we are looking for a percentage increase, then we must have \(a>0\) and so \(a=-1.65\) is inadmissible.
Therefore, the dimensions of the screen must each be increased by \(15\%\) and the numbers of pixels/cm must be increased by \(2\times 15\%=30\%\).
The increased length is \(15\times 1.15=17.25\) cm and the increased width is \(10\times 1.15=11.5\) cm. The increased number of pixels/cm is \(80\times 1.3=104\) pixels/cm.