Let
x ----------> the height of the whole poster
y ----------> the width of the whole poster

We need to minimize the area A=x*y

we know that
(x-4)*(y-2)=722
(y-2)=722/(x-4)
(y)=[722/(x-4)]+2

so
A(x)=x*y--------->A(x)=x*{[722/(x-4)]+2}
Need to minimize this function over x > 4

find the derivative------> A1 (x)
A1(x)=2*[8x²-8x-1428]/[(x-4)²]

for A1(x)=0
8x²-8x-1428=0
using a graph tool
gives x=13.87 in

(y)=[722/(x-4)]+2

y=[2x+714]/[x-4]-----> y=[2*13.87+714]/[13.87-4]-----> y=75.15 in

the answer is
the dimensions of the poster will be
the height of the whole poster is 13.87 in
the width of the whole poster is 75.15 in

6 0

3 years ago

The length of the base of an isosceles triangle is x. The length of a leg is 2x - 2. The perimeter of the triangle is 71. Find

Elanso [62]

Answer:

x = 15

Step-by-step explanation:

x + 2x - 2 + 2x - 2 = 71

Solve for x, you get 15.

3 0

2 years ago

Find the value of the lesser root of x 2-8x+7=0

mariarad [96]

Hi there! The answer is x = 1

${x}^{2} - 8x + 7 = 0$

$(x - 7)(x - 1)$
Rule AB = 0, gives A = 0 or B = 0

$x - 7 = 0 \: \: or \: \: x - 1 = 0 \\ x = 7 \: \: \: \: \: \: \: \: \: or \: \: x = 1$
Therefore, the smallest of both roots is x = 1.

7 0

3 years ago