Answer to Question #112995 in Calculus for cece

Question

Johnny is designing a rectangular poster to contain 32 inches squared of printing with a 4​-in margin at the top and bottom and a 2​-in margin at each side. What overall dimensions will minimize the amount of paper​ used?

Answer 1

Consider the printed area of horizontal length $x$ and vertical length $y$ as shown in the figure below:

Area of printed region as shown above in the figure is,

$$xy=32$$

$$y=\frac{32}{x}$$

Here, the objective is to minimize the total area of the poster$A=(x+4)(y+8)$

Substitute $\frac{32}{x}$ for $y$ into $A=(x+4)(y+8)$ to express the total area in terms of single variable $x$ as,

$$A(x)=(x+4)(\frac{32}{x}+8)$$

Differentiate with respect to $x$ and equate to zero in order to find the critical value as,

$A'(x)=0$

$(x+4)\frac{d}{dx}(\frac{32}{x}+8)+(\frac{32}{x}+8)\frac{d}{dx}(x+4)=0$

$(x+4)(-\frac{32}{x^2})+(\frac{32}{x}+8)(1)=0$

$-\frac{32}{x}-\frac{128}{x^2}+\frac{32}{x}+8=0$

$-\frac{128}{x^2}+8=0$

$x^2=16$

$x=4$

Next plug $x=4$ into $y=\frac{32}{x}$ to obtain $y=\frac{32}{4}=8$

Here, the first derivative change sign from negative to positive at $x=4$ , therefore, there exists a minima at $x=4$ .

Hence, the overall dimensions of poster is:
Horizontal length of poster$=x+4=4+4=8$ in.Vertical length of poster$=y+8=8+8=16$ in.