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Use calculus to find the dimensions of a rectangle with area of 196 square-feet that has the smallest perimeter.

Sagot :

Explanation

In the question, we are given that the area of the rectangle is;

[tex]\text{Area}=196\text{ square fe}et[/tex]

Recall that the area and perimeter of a rectangle are given by the formulas below.

[tex]\begin{gathered} \text{Area = Lenth x Width = L}\times W \\ \text{Perimeter = 2(L+W)} \end{gathered}[/tex]

From the area of the rectangle, we can isolate the variable of the width.

[tex]\begin{gathered} \text{Area}=\text{ L x W} \\ W=\frac{\text{Area}}{L} \\ W=\frac{196}{L} \end{gathered}[/tex]

Therefore, the formula for the perimeter is transformed to give;

[tex]\begin{gathered} \text{Perimeter = 2( L + }\frac{\text{196}}{L}) \\ \text{Simplifying the expression gives;} \\ P=2(\frac{L^2+196}{L}) \\ P=\frac{2L^2+392^{}}{L} \\ P=2L+392L^{-1} \end{gathered}[/tex]

Recall, via the rules of differentiation

[tex]\begin{gathered} \text{for y = x}^n \\ \frac{dy}{dx}=nx^{n-1} \end{gathered}[/tex]

Therefore,

[tex]\begin{gathered} \frac{dP}{dL}=2-392L^{-2}^{} \\ \text{But }\frac{dP}{dL}=0 \\ 0=2-392L^{-2} \\ 392^{}L^{-2}=2 \\ \frac{1}{L^2}=\frac{2}{392}^{} \\ L^2=\frac{392}{2} \\ L^2=196 \\ L=\sqrt[]{196} \\ L=14 \end{gathered}[/tex]

Since

[tex]W=\frac{196}{L}=\frac{196}{14}=14[/tex]

Answer: Length = 14 and Width = 14