If a total of 1100 square centimeters of material is to be used to make a box with a square base and an open top, find the largest possible volume of such a box.

Step-by-step explanation:

Let x = base side length

y = side height

volume V = x²y

y = V/x²

surface area

A = x² + 4xy

A = x² + 4x(V/x²)

A = x² + 4V/x

Solving for volume

V = (A - x²)(x/4)

V = ¼(Ax - x³)

a maximum will occur when the derivative equals zero

V' = ¼(A - 3x²)

0 = ¼(A - 3x²)

0 = A - 3x²

3x² = A

3x² = 1100

x² = (1100 / 3)

x = √(1100/3)

V = (A - x²)(x/4)

V = (1100 - √(1100/3))²(√(1100/3)/4)

V = 3,510.56606...

V = 3510.6 cm³

The box height will be half of the base side length.

y = ½√(1100/3)

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.