Mathematics, 28.10.2020 17:00 thasen31
a box has a square base of side x and height h find the dimensions x, h for which the volume is 14 and the surface area is as small as possible
Answers: 2
![\frac{dS}{dx} = 4x-\frac{56}{x^2}\\0 = 4x-\frac{56}{x^2}\\4x = \frac{56}{x^2}\\4x^3 = 56\\ x^3=14\\x=\sqrt[3]{14}\\x= 2.41](/tpl/images/0847/5687/48512.png)
in above equation, we get :
:![\dfrac{dA}{dx}=2( 2x -\dfrac{28}{x^2})\\\\\dfrac{dA}{dx }= 4(x - \dfrac{14}{x^2})\\\\x - \dfrac{14}{x^2}=0\\\\x=\sqrt[3] {14} = 2.41](/tpl/images/0847/5687/a74a7.png)
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a box has a square base of side x and height h find the dimensions x, h for which the volume is 14 a...
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