Find the dimensions of a box with a square base with surface area 5656 and the maximal volume. (Use symbolic notation and fractions where needed.)

Answer:

28.51 cubic units

Step-by-step explanation:

From the given information;

The surface area is 56 and not 5656

Let assume that:

the length of the base = a &

the height of the box = h

∴

The volume of the box = a²h &

The surface area (A) = 2a² + 4ah

A = 2a² + 4ah

56 = 2a² + 4ah

56 - 2a²  = 4ah

[tex]h = \dfrac{56-2a^2}{4a}[/tex]

The volume (V) = a²h

[tex]V = a^ 2( \dfrac{56-2a^2}{4a}) \\ \\ V = \dfrac{56a -2a^3}{4} \\ \\ V = \dfrac{2a(28- a^2)}{4} \\ \\ V = \dfrac{(28-a^2)a}{2}[/tex]

Taking the maximum of this function by assuming v = [0, [tex]\sqrt{8}[/tex]]

[tex]V = \dfrac{a(28-a^2)}{2}[/tex]

[tex]V(0) = \dfrac{0(28-0^2)}{2}=0[/tex]

[tex]V(\sqrt{8}) = \dfrac{\sqrt{8}(28-(\sqrt{8})^2)}{2} \\ \\ \implies \sqrt{8} \dfrac{(28-8)} {2} \\ \\ = 10 \sqrt{8}[/tex]

For the critical point V' = 0

[tex]V = \dfrac{a(28-a^2)}{2} \\ \\ V = \dfrac{28a}{2}- \dfrac{a^3}{a} \\ \\ V = 14 a - \dfrac{a^3}{2} \\ \\ V' = 14 - \dfrac{3a^2}{2} \\ \\ 0 = 14 - \dfrac{3a^2}{2} \\ \\ 14 =\dfrac{3a^2}{2}\\ \\ a^2 = 14 \times \dfrac{2}{3} \\ \\ a^2 = \dfrac{28}{3} \\ \\ a = \sqrt{\dfrac{28}{3}} \\ \\ a = \sqrt{9.333} \\ \\ a = 3.055[/tex]

Thus, side of the base (a) = 3.055 units

Recall that:

height  [tex]h = \dfrac{56-2a^2}{4a}[/tex]

[tex]h = \dfrac{56-2(3.055)^2}{4(3.055)}[/tex]

h = 3.0551 units

The maximum volume now = a²h

= (3.055)²(3.0551)

= (9.333)(3.0551)

= 28.51 cubic units

Answer:

x ( side of the square base) = 43,42  ul

h = ( the height f the box ) = 0,17 ul

Step-by-step explanation:

V = Ab * h    where   Ab  is an area of the base and h is the height of the box

as is this case s a square base box then  Ab = x²    ( s the side of the square.

The surface area is ( box without lid): area of the base   x²  plus 4 lateral areas each one equal to  x*h

So   x² +  4 x*h = 5656   u²

Then    4*x*h  = 5656 - x²

h  =  5656 - x² )/ 4*x

V(x) = x² ( 5656  -  x²  )/ 4*x

V(x) = x ( 5656  - x² ) / 4

Tacking derivatives on both sides of the equation :

V´(x) = (1/4 ) * [ (5656 - x² ) - x *2*x]

V´(x) = (1/4 ) * ( 5656 - 3*x² )

v¨(x) = 0        (1/4 ) * ( 5656 - 3*x² ) = 0    3*x² = 5656

x² = 1885,33

x  = √ 1885,33

x  =  43,42  ul

And h  = 5656 - x² )/ 4*x

h  =  [( 5656 - (43,42)²] /4 * 5656

h  = 3770,70 / 22624

h  =  0,17 ul

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