What is the y-intercept of the line given by the equation below?

You are planning to make an open rectangular box from aa 4141-in.-by-8181-in. piece of cardboard by cutting congruent squares

marissa [1.9K]

Answer:

13058.83 cubic inches

Step-by-step explanation:

Given that a rectangular box is having dimensions as 41x81 inches.

Let x be the side of square cut from all the four corners.

The open box made would have height as x and length 41-2x with width 81-2x

Volume =

$V(x) = x(41-2x)(81-2x)\\V(x) =x(3321-244x+4x^2)\\V(x) = 3321x-244x^2+4x^3\\V'(x) = 3321-488x+12x^2\\V"(x) = -488+24x\\$

Equate first derivative to 0

We get applicable root as x = 8.642

Max volume = 13058.83 cubic inches

3 0

3 years ago

translate this equation. 126 is the product of goran's score and 7..use the variable g to represent gorans score

FromTheMoon [43]

7g = 126 is the translation

4 0

3 years ago

Read 2 more answers

If f(x)= 3x+5 and g(x) = 3x^2 evaluate f(-8)

Reika [66]

Answer:OOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF

Step-by-step explanation:OOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF

5 0

3 years ago

If p and q vary inversely and p is 10 when q is 19 , determine q when p is equal to 2

GenaCL600 [577]

You divide 10 by 5 is two so 19 decided by five is 3.8

6 0

3 years ago

Read 2 more answers

The top and bottom margins of a poster are each 12 cm and the side margins are each 8 cm. The area of printed material on the po

grin007 [14]

Answer:

Dimensions of printed poster are

length is 32 cm

width is 48 cm

Step-by-step explanation:

Let's assume

length of printed poster is x cm

width of printed poster is y cm

now, we can find area of printed poster

so, area of printed poster is

$=xy$

we are given that area as 1536

so, we can set it to 1536

$xy=1536$

now, we can solve for y

$y=\frac{1536}{x}$

now, we are given

The top and bottom margins of a poster are each 12 cm and the side margins are each 8 cm

so, total area of poster is

$A=(8+x+8)\times (12+y+12)$

$A=(x+16)\times (y+24)$

now, we can plug back y

$A=(x+16)\times (\frac{1536}{x}+24)$

now, we have to minimize A

so, we will find derivative

$A'=\frac{d}{dx}\left(\left(x+16\right)\left(\frac{1536}{x}+24\right)\right)$

we can use product rule

$A'=\frac{d}{dx}\left(x+16\right)\left(\frac{1536}{x}+24\right)+\frac{d}{dx}\left(\frac{1536}{x}+24\right)\left(x+16\right)$

$=\frac{d}{dx}\left(x+16\right)\left(\frac{1536}{x}+24\right)+\frac{d}{dx}\left(\frac{1536}{x}+24\right)\left(x+16\right)$

now, we can simplify it

$A'=-\frac{24576}{x^2}+24$

now, we can set it to 0

and then we can solve for x

$A'=-\frac{24576}{x^2}+24=0$

$-\frac{24576}{x^2}x^2+24x^2=0\cdot \:x^2$

$-24576+24x^2=0$

$x=32,\:x=-32$

Since, x is dimension

and dimension can never be negative

so, we will only consider positive value

$x=32$

now, we can solve for y

$y=\frac{1536}{32}$

$y=48$

so, dimensions of printed poster are

length is 32 cm

width is 48 cm

5 0

3 years ago