3a+2b=193 Find a and b

If 1200 cm2 of material is available to make a box with a square base and an open top, find the largest possible volume of the b

Colt1911 [192]

Answer:

The largest possible volume V is ;

V = l^2 × h

V = 20^2 × 10 = 4000cm^3

Step-by-step explanation:

Given

Volume of a box = length × breadth × height= l×b×h

In this case the box have a square base. i.e l=b

Volume V = l^2 × h

The surface area of a square box

S = 2(lb+lh+bh)

S = 2(l^2 + lh + lh) since l=b

S = 2(l^2 + 2lh)

Given that the box is open top.

S = l^2 + 4lh

And Surface Area of the box is 1200cm^2

1200 = l^2 + 4lh ....1

Making h the subject of formula

h = (1200 - l^2)/4l .....2

Volume is given as

V = l^2 × h

V = l^2 ×(1200 - l^2)/4l

V = (1200l - l^3)/4

the maximum point is at dV/dl = 0

dV/dl = (1200 - 3l^2)/4

dV/dl = (1200 - 3l^2)/4 = 0

3l^2= 1200

l^2 = 1200/3 = 400

l = √400

I = 20cm

Since,

h = (1200 - l^2)/4l

h = (1200 - 20^2)/4×20

h = (800)/80

h = 10cm

The largest possible volume V is ;

V = l^2 × h

V = 20^2 × 10 = 4000cm^3

4 0

3 years ago

HELPPPPP!!!! ASAPPPPP!!!!!

kaheart [24]

Answer:

24

Step-by-step explanation:

XW=WY=12

If the opposite angle is 30, the hypotenuse is double the small side.

XZ=XW=2·12=24

5 0

3 years ago

Simplify the expression and identify which property is used in each step.

777dan777 [17]

Answer:

1.2

Step-by-step explanation:

(0.85 + 0.50 + 0.15) - 0.30 Given

Simplify by adding in the parentheses first

(0.85 + 0.50 + 0.15) - 0.30

(1.35 + 0.15) - 0.30

(1.5) - 0.30

Subtraction Property

1.5 - 0.30

1.2

7 0

3 years ago

Factorize ху + 3x +3y + 18

noname [10]

Answer:

(x+6)(y+3)

Step-by-step explanation:

xy + 3x + 6y + 18
x(y + 3) + 6(y+3)
(x+6)(y+3)

3 0

2 years ago

Write the congruency statement for the triangles. _______In triangles MNO and JKL, there are two congruent sides and included an

mafiozo [28]

Answer:

ΔJKL ≅ ΔMNO by SAS

Step-by-step explanation:

The question is incomplete, the question is as following:

ΔMNO is shown.

Which Δ below can be shown to be congruent to ΔMNO with only the given information?

Name the postulate that justifies your answer (SAS, AAS, HL, ASA or SSS).

Write the congruency statement for the triangles.

And see the attached figure which represent ΔMNO

==========================================================

Check the given options to see which triangle from the option will be congruent to ΔMNO

1. In triangles MNO and ABC, there are two congruent sides and non-included angle - which mean (angle - side - side)

The (angle - side - side) Postulate does not exist because an angle and two sides does not guarantee that two triangles are congruent.

2. In triangles MNO and DEF, there are two congruent sides - there is not enough information

3. In triangles MNO and GHI, there are three congruent angles - which mean AAA (angle - angle - angle)

The AAA Postulate does not exist because an angle and two sides does not guarantee that two triangles are congruent. This postulate is used to prove the similarity between the triangles.

4. In triangles MNO and JKL, there are two congruent sides and included angle - SAS (side - angle - side)

MN ≅ JK , MO ≅ JL , ∠M ≅ ∠J

5 0

3 years ago