Chapter ISG, Problem 9.6.7P

a.

To determine

To find dimensions of rectangle.

Answer to Problem 9.6.7P

No width

Explanation of Solution

Given information:

Length of rectangle is $15-w$

Width of rectangle is $w$

Area of rectangle is $A=150i{n}^2$

Formula used:

  $A=lw$

Where

Length of rectangle is l

Width of rectangle is $w$

Calculation:

Length of rectangle $=15-w$

Width of rectangle $=w$

Area of rectangle, $A=150i{n}^2$

Using formula

  $A=(15-w)w$

put value of area

  $150=(15-w)w$

On solving,

  $150=15w-w^2$

On solving,

  $w^2-15w+150=0$

On solving,

  $w=\frac{-(-15)\pm \sqrt{{(-15)}^2-4(1)(150)}}{2(1)}$

On solving,

  $w=\frac{15\pm \sqrt{225-600}}{2}$

On solving,

  $w=\frac{15\pm \sqrt{-375}}{2}$

The expression is undefined in the set of real numbers

Hence, no width

b.

To determine

To find maximum dimensions and maximum area of rectangle.

Answer to Problem 9.6.7P

Maximum length of rectangle $=\frac{15}{2}$

Maximum width of rectangle $=\frac{15}{2}$

Maximum area of rectangle

  $A=56.25i{n}^2$

Explanation of Solution

Given information:

Length of rectangle is $15-w$

Width of rectangle is $w$

Area of rectangle is $A=150i{n}^2$

Formula used:

  $A=lw$

Where

Length of rectangle is l

Width of rectangle is $w$

Maximum length, $l=\frac{perimeter}{4}$

Maximum width, $w=\frac{perimeter}{4}$

Calculation:

Maximum length of rectangle $=\frac{15}{2}$

Maximum width of rectangle $=\frac{15}{2}$

Maximum area of rectangle, Using formula

  $A=\frac{15}{2}\times \frac{15}{2}$

On solving,

  $A=\frac{225}{4}$

On solving,

  $A=56.25i{n}^2$